def test_dtypes_of_operator_sum():
# gh-6078
mat_complex = np.random.rand(2, 2) + 1j * np.random.rand(2, 2)
mat_real = np.random.rand(2, 2)
complex_operator = interface.aslinearoperator(mat_complex)
real_operator = interface.aslinearoperator(mat_real)
sum_complex = complex_operator + complex_operator
sum_real = real_operator + real_operator
assert_equal(sum_real.dtype, np.float64)
assert_equal(sum_complex.dtype, np.complex128)
def test_dtypes_of_operator_sum():
# gh-6078
mat_complex = np.random.rand(2,2) + 1j * np.random.rand(2,2)
mat_real = np.random.rand(2,2)
complex_operator = interface.aslinearoperator(mat_complex)
real_operator = interface.aslinearoperator(mat_real)
sum_complex = complex_operator + complex_operator
sum_real = real_operator + real_operator
assert_equal(sum_real.dtype, np.float64)
assert_equal(sum_complex.dtype, np.complex128)
def setUp(self):
self.n = 10
self.A = lowerBidiagonalMatrix(20,self.n)
self.xtrue = transpose(arange(self.n,0,-1))
self.Afun = aslinearoperator(self.A)
self.b = self.Afun.matvec(self.xtrue)
self.returnValues = lsmr(self.A,self.b)
def testComplexX0(self):
A = 4 * eye(self.n) + ones((self.n, self.n))
xtrue = transpose(arange(self.n, 0, -1))
b = aslinearoperator(A).matvec(xtrue)
x0 = zeros(self.n, dtype=complex)
x = lsmr(A, b, x0=x0)[0]
assert norm(x - xtrue) == pytest.approx(0, abs=1e-5)
def test_basic(self):
for M in self.cases:
A = interface.aslinearoperator(M)
M, N = A.shape
assert_equal(A.matvec(np.array([1, 2, 3])), [14, 32])
assert_equal(A.matvec(np.array([[1], [2], [3]])), [[14], [32]])
assert_equal(A * np.array([1, 2, 3]), [14, 32])
assert_equal(A * np.array([[1], [2], [3]]), [[14], [32]])
assert_equal(A.rmatvec(np.array([1, 2])), [9, 12, 15])
assert_equal(A.rmatvec(np.array([[1], [2]])), [[9], [12], [15]])
assert_equal(A.H.matvec(np.array([1, 2])), [9, 12, 15])
assert_equal(A.H.matvec(np.array([[1], [2]])), [[9], [12], [15]])
assert_equal(A.matmat(np.array([[1, 4], [2, 5], [3, 6]])),
[[14, 32], [32, 77]])
assert_equal(A * np.array([[1, 4], [2, 5], [3, 6]]),
[[14, 32], [32, 77]])
if hasattr(M, 'dtype'):
assert_equal(A.dtype, M.dtype)
def testComplexX0(self):
A = 4 * eye(self.n) + ones((self.n, self.n))
xtrue = transpose(arange(self.n, 0, -1))
b = aslinearoperator(A).matvec(xtrue)
x0 = zeros(self.n, dtype=complex)
x = lsmr(A, b, x0=x0)[0]
assert_almost_equal(norm(x - xtrue), 0, decimal=5)
def test_basic(self):
for M in self.cases:
A = interface.aslinearoperator(M)
M,N = A.shape
assert_equal(A.matvec(np.array([1,2,3])), [14,32])
assert_equal(A.matvec(np.array([[1],[2],[3]])), [[14],[32]])
assert_equal(A * np.array([1,2,3]), [14,32])
assert_equal(A * np.array([[1],[2],[3]]), [[14],[32]])
assert_equal(A.rmatvec(np.array([1,2])), [9,12,15])
assert_equal(A.rmatvec(np.array([[1],[2]])), [[9],[12],[15]])
assert_equal(A.H.matvec(np.array([1,2])), [9,12,15])
assert_equal(A.H.matvec(np.array([[1],[2]])), [[9],[12],[15]])
assert_equal(
A.matmat(np.array([[1,4],[2,5],[3,6]])),
[[14,32],[32,77]])
assert_equal(A * np.array([[1,4],[2,5],[3,6]]), [[14,32],[32,77]])
if hasattr(M,'dtype'):
assert_equal(A.dtype, M.dtype)
def test_basic(self):
for M, A_array in self.cases:
A = interface.aslinearoperator(M)
M,N = A.shape
xs = [np.array([1, 2, 3]),
np.array([[1], [2], [3]])]
ys = [np.array([1, 2]), np.array([[1], [2]])]
if A.dtype == np.complex_:
xs += [np.array([1, 2j, 3j]),
np.array([[1], [2j], [3j]])]
ys += [np.array([1, 2j]), np.array([[1], [2j]])]
x2 = np.array([[1, 4], [2, 5], [3, 6]])
for x in xs:
assert_equal(A.matvec(x), A_array.dot(x))
assert_equal(A * x, A_array.dot(x))
assert_equal(A.matmat(x2), A_array.dot(x2))
assert_equal(A * x2, A_array.dot(x2))
for y in ys:
assert_equal(A.rmatvec(y), A_array.T.conj().dot(y))
assert_equal(A.T.matvec(y), A_array.T.dot(y))
assert_equal(A.H.matvec(y), A_array.T.conj().dot(y))
if hasattr(M,'dtype'):
assert_equal(A.dtype, M.dtype)
assert_(hasattr(A, 'args'))
def test(self):
maxn = 15 # Dimension of square matrix to be solved
# Use a PDP^-1 factorisation to construct matrix with known
# eiegevalues/vectors. Used random eiegenvectors initially.
P = np.mat(np.random.random((maxn, ) * 2))
P /= map(np.linalg.norm, P.T) # Normalise the eigenvectors
D = np.mat(np.zeros((maxn, ) * 2))
D[range(maxn),
range(maxn)] = (np.arange(maxn, dtype=float) + 1) / np.sqrt(maxn)
A = P * D * np.linalg.inv(P)
vals = np.array(D.diagonal())[0]
vecs = P
uv_sortind = vals.argsort()
vals = vals[uv_sortind]
vecs = vecs[:, uv_sortind]
A = aslinearoperator(A)
matvec = A.matvec
#= lambda x: np.asarray(A*x)[0]
nev = 4
eigvs = ARPACK_eigs(matvec, A.shape[0], nev=nev)
calc_vals = eigvs[0]
# Ensure the calculated eigenvectors have the same sign as the reference values
calc_vecs = eigvs[1] / [np.sign(x[0]) for x in eigvs[1].T]
assert_array_almost_equal(calc_vals, vals[0:nev], decimal=7)
assert_array_almost_equal(calc_vecs,
np.array(vecs)[:, 0:nev],
decimal=7)
def test(self):
maxn=15 # Dimension of square matrix to be solved
# Use a PDP^-1 factorisation to construct matrix with known
# eiegevalues/vectors. Used random eiegenvectors initially.
P = np.mat(np.random.random((maxn,)*2))
P /= map(np.linalg.norm, P.T) # Normalise the eigenvectors
D = np.mat(np.zeros((maxn,)*2))
D[range(maxn), range(maxn)] = (np.arange(maxn, dtype=float)+1)/np.sqrt(maxn)
A = P*D*np.linalg.inv(P)
vals = np.array(D.diagonal())[0]
vecs = P
uv_sortind = vals.argsort()
vals = vals[uv_sortind]
vecs = vecs[:,uv_sortind]
A=aslinearoperator(A)
matvec = A.matvec
#= lambda x: np.asarray(A*x)[0]
nev=4
eigvs = ARPACK_eigs(matvec, A.shape[0], nev=nev)
calc_vals = eigvs[0]
# Ensure the calculated eigenvectors have the same sign as the reference values
calc_vecs = eigvs[1] / [np.sign(x[0]) for x in eigvs[1].T]
assert_array_almost_equal(calc_vals, vals[0:nev], decimal=7)
assert_array_almost_equal(calc_vecs, np.array(vecs)[:,0:nev], decimal=7)
def setUp(self):
self.n = 10
self.A = lowerBidiagonalMatrix(20,self.n)
self.xtrue = transpose(arange(self.n,0,-1))
self.Afun = aslinearoperator(self.A)
self.b = self.Afun.matvec(self.xtrue)
self.returnValues = lsmr(self.A,self.b)
def setup_method(self):
self.n = 10
self.A = lowerBidiagonalMatrix(20, self.n)
self.xtrue = transpose(arange(self.n, 0, -1))
self.Afun = aslinearoperator(self.A)
self.b = self.Afun.matvec(self.xtrue)
self.x0 = ones(self.n)
self.x00 = self.x0.copy()
self.returnValues = lsmr(self.A, self.b)
self.returnValuesX0 = lsmr(self.A, self.b, x0=self.x0)
def test_dot(self):
for M in self.cases:
A = interface.aslinearoperator(M)
M, N = A.shape
assert_equal(A.dot(np.array([1, 2, 3])), [14, 32])
assert_equal(A.dot(np.array([[1], [2], [3]])), [[14], [32]])
assert_equal(A.dot(np.array([[1, 4], [2, 5], [3, 6]])), [[14, 32], [32, 77]])
def test_adjoint_conjugate():
X = np.array([[1j]])
A = interface.aslinearoperator(X)
B = 1j * A
Y = 1j * X
v = np.array([1])
assert_equal(B.dot(v), Y.dot(v))
assert_equal(B.H.dot(v), Y.T.conj().dot(v))
def test_adjoint_conjugate():
X = np.array([[1j]])
A = interface.aslinearoperator(X)
B = 1j * A
Y = 1j * X
v = np.array([1])
assert_equal(B.dot(v), Y.dot(v))
assert_equal(B.H.dot(v), Y.T.conj().dot(v))
def test_dot(self):
for M in self.cases:
A = interface.aslinearoperator(M)
M, N = A.shape
assert_equal(A.dot(np.array([1, 2, 3])), [14, 32])
assert_equal(A.dot(np.array([[1], [2], [3]])), [[14], [32]])
assert_equal(A.dot(np.array([[1, 4], [2, 5], [3, 6]])),
[[14, 32], [32, 77]])
def test_transpose_noconjugate():
X = np.array([[1j]])
A = interface.aslinearoperator(X)
B = 1j * A
Y = 1j * X
v = np.array([1])
assert_equal(B.dot(v), Y.dot(v))
assert_equal(B.T.dot(v), Y.T.dot(v))
def test_dot(self):
for M, A_array in self.cases:
A = interface.aslinearoperator(M)
M, N = A.shape
x0 = np.array([1, 2, 3])
x1 = np.array([[1], [2], [3]])
x2 = np.array([[1, 4], [2, 5], [3, 6]])
assert_equal(A.dot(x0), A_array.dot(x0))
assert_equal(A.dot(x1), A_array.dot(x1))
assert_equal(A.dot(x2), A_array.dot(x2))
def test_attributes():
A = interface.aslinearoperator(np.arange(16).reshape(4, 4))
def always_four_ones(x):
x = np.asarray(x)
assert_(x.shape == (3,) or x.shape == (3, 1))
return np.ones(4)
B = interface.LinearOperator(shape=(4, 3), matvec=always_four_ones)
for op in [A, B, A * B, A.H, A + A, B + B, A ** 4]:
assert_(hasattr(op, "dtype"))
assert_(hasattr(op, "shape"))
assert_(hasattr(op, "_matvec"))
def test_attributes():
A = interface.aslinearoperator(np.arange(16).reshape(4, 4))
def always_four_ones(x):
x = np.asarray(x)
assert_(x.shape == (3, ) or x.shape == (3, 1))
return np.ones(4)
B = interface.LinearOperator(shape=(4, 3), matvec=always_four_ones)
for op in [A, B, A * B, A.H, A + A, B + B, A**4]:
assert_(hasattr(op, "dtype"))
assert_(hasattr(op, "shape"))
assert_(hasattr(op, "_matvec"))
def arpack_eigs(mtx, nev=1, which='SM'):
"""
Calculate several eigenvalues and corresponding eigenvectors of a
matrix using ARPACK from SciPy. The eigenvalues are sorted in
ascending order.
"""
from scipy.sparse.linalg.interface import aslinearoperator
from scipy.sparse.linalg.eigen.arpack import speigs
matvec = aslinearoperator(mtx).matvec
eigs, vecs = speigs.ARPACK_eigs(matvec, mtx.shape[0], nev=nev, which=which)
ii = nm.argsort(eigs)
eigs = eigs[ii]
vecs = vecs[:,ii]
return eigs, vecs
def lsmrtest(m, n, damp):
"""Verbose testing of lsmr"""
A = lowerBidiagonalMatrix(m,n)
xtrue = arange(n,0,-1, dtype=float)
Afun = aslinearoperator(A)
b = Afun.matvec(xtrue)
atol = 1.0e-7
btol = 1.0e-7
conlim = 1.0e+10
itnlim = 10*n
show = 1
x, istop, itn, normr, normar, norma, conda, normx \
= lsmr(A, b, damp, atol, btol, conlim, itnlim, show)
j1 = min(n,5)
j2 = max(n-4,1)
print(' ')
print('First elements of x:')
str = ['%10.4f' % (xi) for xi in x[0:j1]]
print(''.join(str))
print(' ')
print('Last elements of x:')
str = ['%10.4f' % (xi) for xi in x[j2-1:]]
print(''.join(str))
r = b - Afun.matvec(x)
r2 = sqrt(norm(r)**2 + (damp*norm(x))**2)
print(' ')
str = 'normr (est.) %17.10e' % (normr)
str2 = 'normr (true) %17.10e' % (r2)
print(str)
print(str2)
print(' ')
def svd(A, k=6):
"""Compute a few singular values/vectors for a sparse matrix using ARPACK.
Parameters
----------
A: sparse matrix
Array to compute the SVD on.
k: int
Number of singular values and vectors to compute.
Note
----
This is a naive implementation using the symmetric eigensolver on A.T * A
or A * A.T, depending on which one is more efficient.
Complex support is not implemented yet
"""
# TODO: implement complex support once ARPACK-based eigen_hermitian is
# available
n, m = A.shape
if np.iscomplexobj(A):
raise NotImplementedError("Complex support for sparse SVD not "
"implemented yet")
op = lambda x: x.T.conjugate()
else:
op = lambda x: x.T
tp = A.dtype.char
linear_at = aslinearoperator(op(A))
linear_a = aslinearoperator(A)
def _left(x, sz):
x = csc_matrix(x)
matvec = lambda x: linear_at.matvec(linear_a.matvec(x))
params = _SymmetricArpackParams(sz, k, tp, matvec)
while not params.converged:
params.iterate()
eigvals, eigvec = params.extract(True)
s = np.sqrt(eigvals)
v = eigvec
u = (x * v) / s
return u, s, op(v)
def _right(x, sz):
x = csr_matrix(x)
matvec = lambda x: linear_a.matvec(linear_at.matvec(x))
params = _SymmetricArpackParams(sz, k, tp, matvec)
while not params.converged:
params.iterate()
eigvals, eigvec = params.extract(True)
s = np.sqrt(eigvals)
u = eigvec
vh = (op(u) * x) / s[:, None]
return u, s, vh
if n > m:
return _left(A, m)
else:
return _right(A, n)
def assertCompatibleSystem(self, A, xtrue):
Afun = aslinearoperator(A)
b = Afun.matvec(xtrue)
x = lsmr(A, b)[0]
assert norm(x - xtrue) == pytest.approx(0, abs=1e-5)
def eigsh(A,
k=6,
M=None,
sigma=None,
which='LM',
v0=None,
ncv=None,
maxiter=None,
tol=0,
return_eigenvectors=True,
Minv=None,
OPinv=None,
mode='normal',
lock=None,
return_stats=False,
maxBlockSize=0,
minRestartSize=0,
maxPrevRetain=0,
method=None):
"""
Find k eigenvalues and eigenvectors of the real symmetric square matrix
or complex Hermitian matrix A.
Solves ``A * x[i] = w[i] * x[i]``, the standard eigenvalue problem for
w[i] eigenvalues with corresponding eigenvectors x[i].
If M is specified, solves ``A * x[i] = w[i] * M * x[i]``, the
generalized eigenvalue problem for w[i] eigenvalues
with corresponding eigenvectors x[i]
Parameters
----------
A : An N x N matrix, array, sparse matrix, or LinearOperator representing
the operation A * x, where A is a real symmetric matrix or complex
Hermitian.
k : int, optional
The number of eigenvalues and eigenvectors desired.
`k` must be smaller than N. It is not possible to compute all
eigenvectors of a matrix.
M : An N x N matrix, array, sparse matrix, or LinearOperator representing
the operation M * x for the generalized eigenvalue problem
A * x = w * M * x.
M must represent a real, symmetric matrix if A is real, and must
represent a complex, hermitian matrix if A is complex. For best
results, the data type of M should be the same as that of A.
sigma : real, optional
Find eigenvalues near sigma.
v0 : N x i, ndarray, optional
Starting vectors for iteration.
ncv : int, optional
The maximum size of the basis
which : str ['LM' | 'SM' | 'LA' | 'SA' | 'BE']
If A is a complex hermitian matrix, 'BE' is invalid.
Which `k` eigenvectors and eigenvalues to find:
'LM' : Largest (in magnitude) eigenvalues
'SM' : Smallest (in magnitude) eigenvalues
'LA' : Largest (algebraic) eigenvalues
'SA' : Smallest (algebraic) eigenvalues
'BE' : Half (k/2) from each end of the spectrum (not supported)
When sigma != None, 'which' refers to the shifted eigenvalues ``w'[i]``
maxiter : int, optional
Maximum number of iterations.
tol : float
Accuracy for eigenvalues (stopping criterion).
The default value is sqrt of machine precision.
Minv : (not supported)
OPinv : N x N matrix, array, sparse matrix, or LinearOperator
Preconditioner to accelerate the convergence. Usually it is an
approximation of the inverse of (A - sigma*M).
return_eigenvectors : bool
Return eigenvectors (True) in addition to eigenvalues
mode : string ['normal' | 'buckling' | 'cayley']
Only 'normal' mode is supported.
lock : N x i, ndarray, optional
Seek the eigenvectors orthogonal to these ones. The provided
vectors *should* be orthonormal. Useful to not converge some already
computed solutions.
maxBlockSize : int, optional
Maximum number of vectors added at every iteration.
minRestartSize : int, optional
Number of approximate eigenvectors kept from last iteration in restart.
maxPrevRetain: int, optional
Number of approximate eigenvectors kept from previous iteration in
restart. Also referred as +k vectors in GD+k.
method : int, optional
Preset method, one of:
- DEFAULT_MIN_TIME : a variant of JDQMR,
- DEFAULT_MIN_MATVECS : GD+k
- DYNAMIC : choose dynamically between both previous methods.
See a detailed description of the methods and other possible values
in [2]_.
report_stats : bool, optional
If True, it is also returned extra information from PRIMME.
Returns
-------
w : array
Array of k eigenvalues
v : array
An array representing the `k` eigenvectors. The column ``v[:, i]`` is
the eigenvector corresponding to the eigenvalue ``w[i]``.
stats : dict, optional (if return_stats)
Extra information reported by PRIMME:
- "numOuterIterations": number of outer iterations
- "numRestarts": number of restarts
- "numMatvecs": number of A*v
- "numPreconds": number of OPinv*v
- "elapsedTime": time that took
- "estimateMinEVal": the leftmost Ritz value seen
- "estimateMaxEVal": the rightmost Ritz value seen
- "estimateLargestSVal": the largest singular value seen
Raises
------
PrimmeError
When the requested convergence is not obtained.
The PRIMME error code can be found as ``err`` attribute of the exception
object.
See Also
--------
scipy.sparse.linalg.eigs : eigenvalues and eigenvectors for a general (nonsymmetric) matrix A
Primme.svds : singular value decomposition for a matrix A
Notes
-----
This function is a wrapper to PRIMME functions to find the eigenvalues and
eigenvectors [1]_.
References
----------
.. [1] PRIMME Software, https://github.com/primme/primme
.. [2] Preset Methods, http://www.cs.wm.edu/~andreas/software/doc/readme.html#preset-methods
Examples
--------
>>> import Primme, scipy.sparse
>>> A = scipy.sparse.spdiags(range(100), [0], 100, 100) # sparse diag. matrix
>>> evals, evecs = Primme.eigsh(A, 3, tol=1e-6, which='LA')
>>> evals # the three largest eigenvalues of A
array([ 99., 98., 97.])
>>> evals, evecs = Primme.eigsh(A, 3, tol=1e-6, which='LA', lock=evecs)
>>> evals # the next three largest eigenvalues
array([ 96., 95., 94.])
"""
A = aslinearoperator(A)
if len(A.shape) != 2 or A.shape[0] != A.shape[1]:
raise ValueError('A: expected square matrix (shape=%s)' % (A.shape, ))
if M is not None:
raise ValueError('generalized problems (M != None) are not supported')
class PP(PrimmeParams):
def __init__(self):
PrimmeParams.__init__(self)
def matvec(self, X):
return A.matmat(X)
def prevec(self, X):
return OPinv.matmat(X)
pp = PP()
pp.n = A.shape[0]
if k <= 0 or k > pp.n:
raise ValueError("k=%d must be between 1 and %d, the order of the "
"square input matrix." % (k, pp.n))
pp.numEvals = k
pp.correctionParams.precondition = 0 if OPinv is None else 1
if which == 'LM':
pp.target = primme_largest_abs
if sigma is None:
sigma = 0.0
elif which == 'LA':
pp.target = primme_largest
sigma = None
elif which == 'SA':
pp.target = primme_smallest
sigma = None
elif which == 'SM':
pp.target = primme_closest_abs
if sigma is None:
sigma = 0.0
else:
raise ValueError("which='%s' not supported" % which)
if sigma is not None:
pp.targetShifts = np.array([sigma], dtype=np.dtype('d'))
pp.eps = tol
if ncv is not None:
pp.maxBasisSize = ncv
if maxiter is not None:
pp.maxMatvecs = maxiter
if OPinv is not None:
OPinv = aslinearoperator(OPinv)
if OPinv.shape[0] != OPinv.shape[1] or OPinv.shape[0] != A.shape[0]:
raise ValueError(
'OPinv: expected square matrix with same shape as A (shape=%s)'
% (OPinv.shape, ))
pp.correctionParams.precondition = 1
if lock is not None:
if lock.shape[0] != pp.n:
raise ValueError(
'lock: expected matrix with the same columns as A (shape=%s)' %
(lock.shape, ))
pp.numOrthoConst = min(lock.shape[1], pp.n)
if A.dtype.kind in frozenset(["b", "i", "u", "f"]):
dtype = np.dtype("d")
else:
dtype = np.dtype("complex128")
evals = np.zeros(pp.numEvals)
norms = np.zeros(pp.numEvals)
evecs = np.zeros((pp.n, pp.numOrthoConst + pp.numEvals), dtype, order='F')
if lock is not None:
np.copyto(evecs[:, 0:pp.numOrthoConst], lock[:, 0:pp.numOrthoConst])
if v0 is not None:
pp.initSize = min(v0.shape[1], pp.numEvals)
np.copyto(evecs[:, pp.numOrthoConst:pp.numOrthoConst + pp.initSize],
v0[:, 0:pp.initSize])
if maxBlockSize:
pp.maxBlockSize = maxBlockSize
if minRestartSize:
pp.minRestartSize = minRestartSize
if maxPrevRetain:
pp.restartingParams.maxPrevRetain = maxPrevRetain
if method is not None:
pp.set_method(method)
if dtype is np.dtype(np.complex128):
err = zprimme(evals, evecs, norms, pp)
else:
err = dprimme(evals, evecs, norms, pp)
if err != 0:
raise PrimmeError(err)
evecs = evecs[:, pp.numOrthoConst:]
if return_stats:
stats = dict((f, getattr(pp.stats, f)) for f in [
"numOuterIterations", "numRestarts", "numMatvecs", "numPreconds",
"elapsedTime", "estimateMinEVal", "estimateMaxEVal",
"estimateLargestSVal"
])
return evals, evecs, stats
else:
return evals, evecs
def test_ndim():
X = np.array([[1]])
A = interface.aslinearoperator(X)
assert_equal(A.ndim, 2)
def eigen(A, k=6, M=None, sigma=None, which='LM', v0=None,
ncv=None, maxiter=None, tol=0,
return_eigenvectors=True):
"""Find k eigenvalues and eigenvectors of the square matrix A.
Solves A * x[i] = w[i] * x[i], the standard eigenvalue problem for
w[i] eigenvalues with corresponding eigenvectors x[i].
Parameters
----------
A : matrix, array, or object with matvec(x) method
An N x N matrix, array, or an object with matvec(x) method to perform
the matrix vector product A * x. The sparse matrix formats
in scipy.sparse are appropriate for A.
k : integer
The number of eigenvalues and eigenvectors desired
Returns
-------
w : array
Array of k eigenvalues
v : array
An array of k eigenvectors
The v[i] is the eigenvector corresponding to the eigenvector w[i]
Other Parameters
----------------
M : matrix or array
(Not implemented)
A symmetric positive-definite matrix for the generalized
eigenvalue problem A * x = w * M * x
sigma : real or complex
(Not implemented)
Find eigenvalues near sigma. Shift spectrum by sigma.
v0 : array
Starting vector for iteration.
ncv : integer
The number of Lanczos vectors generated
ncv must be greater than k; it is recommended that ncv > 2*k
which : string
Which k eigenvectors and eigenvalues to find:
- 'LM' : largest magnitude
- 'SM' : smallest magnitude
- 'LR' : largest real part
- 'SR' : smallest real part
- 'LI' : largest imaginary part
- 'SI' : smallest imaginary part
maxiter : integer
Maximum number of Arnoldi update iterations allowed
tol : float
Relative accuracy for eigenvalues (stopping criterion)
return_eigenvectors : boolean
Return eigenvectors (True) in addition to eigenvalues
See Also
--------
eigen_symmetric : eigenvalues and eigenvectors for symmetric matrix A
Notes
-----
Examples
--------
"""
A = aslinearoperator(A)
if A.shape[0] != A.shape[1]:
raise ValueError('expected square matrix (shape=%s)' % A.shape)
n = A.shape[0]
# guess type
typ = A.dtype.char
if typ not in 'fdFD':
raise ValueError("matrix type must be 'f', 'd', 'F', or 'D'")
if M is not None:
raise NotImplementedError("generalized eigenproblem not supported yet")
if sigma is not None:
raise NotImplementedError("shifted eigenproblem not supported yet")
# some defaults
if ncv is None:
ncv=2*k+1
ncv=min(ncv,n)
if maxiter==None:
maxiter=n*10
# assign starting vector
if v0 is not None:
resid=v0
info=1
else:
resid = np.zeros(n,typ)
info=0
# some sanity checks
if k <= 0:
raise ValueError("k must be positive, k=%d"%k)
if k == n:
raise ValueError("k must be less than rank(A), k=%d"%k)
if maxiter <= 0:
raise ValueError("maxiter must be positive, maxiter=%d"%maxiter)
whiches=['LM','SM','LR','SR','LI','SI']
if which not in whiches:
raise ValueError("which must be one of %s"%' '.join(whiches))
if ncv > n or ncv < k:
raise ValueError("ncv must be k<=ncv<=n, ncv=%s"%ncv)
# assign solver and postprocessor
ltr = _type_conv[typ]
eigsolver = _arpack.__dict__[ltr+'naupd']
eigextract = _arpack.__dict__[ltr+'neupd']
v = np.zeros((n,ncv),typ) # holds Ritz vectors
workd = np.zeros(3*n,typ) # workspace
workl = np.zeros(3*ncv*ncv+6*ncv,typ) # workspace
iparam = np.zeros(11,'int') # problem parameters
ipntr = np.zeros(14,'int') # pointers into workspaces
ido = 0
if typ in 'FD':
rwork = np.zeros(ncv,typ.lower())
# set solver mode and parameters
# only supported mode is 1: Ax=lx
ishfts = 1
mode1 = 1
bmat = 'I'
iparam[0] = ishfts
iparam[2] = maxiter
iparam[6] = mode1
while True:
if typ in 'fd':
ido,resid,v,iparam,ipntr,info =\
eigsolver(ido,bmat,which,k,tol,resid,v,iparam,ipntr,
workd,workl,info)
else:
ido,resid,v,iparam,ipntr,info =\
eigsolver(ido,bmat,which,k,tol,resid,v,iparam,ipntr,
workd,workl,rwork,info)
xslice = slice(ipntr[0]-1, ipntr[0]-1+n)
yslice = slice(ipntr[1]-1, ipntr[1]-1+n)
if ido == -1:
# initialization
workd[yslice]=A.matvec(workd[xslice])
elif ido == 1:
# compute y=Ax
workd[yslice]=A.matvec(workd[xslice])
else:
break
if info < -1 :
raise RuntimeError("Error info=%d in arpack"%info)
return None
if info == -1:
warnings.warn("Maximum number of iterations taken: %s"%iparam[2])
# if iparam[3] != k:
# warnings.warn("Only %s eigenvalues converged"%iparam[3])
# now extract eigenvalues and (optionally) eigenvectors
rvec = return_eigenvectors
ierr = 0
howmny = 'A' # return all eigenvectors
sselect = np.zeros(ncv,'int') # unused
sigmai = 0.0 # no shifts, not implemented
sigmar = 0.0 # no shifts, not implemented
workev = np.zeros(3*ncv,typ)
if typ in 'fd':
dr=np.zeros(k+1,typ)
di=np.zeros(k+1,typ)
zr=np.zeros((n,k+1),typ)
dr,di,zr,info=\
eigextract(rvec,howmny,sselect,sigmar,sigmai,workev,
bmat,which,k,tol,resid,v,iparam,ipntr,
workd,workl,info)
# The ARPACK nonsymmetric real and double interface (s,d)naupd return
# eigenvalues and eigenvectors in real (float,double) arrays.
# Build complex eigenvalues from real and imaginary parts
d=dr+1.0j*di
# Arrange the eigenvectors: complex eigenvectors are stored as
# real,imaginary in consecutive columns
z=zr.astype(typ.upper())
eps=np.finfo(typ).eps
i=0
while i<=k:
# check if complex
if abs(d[i].imag)>eps:
# assume this is a complex conjugate pair with eigenvalues
# in consecutive columns
z[:,i]=zr[:,i]+1.0j*zr[:,i+1]
z[:,i+1]=z[:,i].conjugate()
i+=1
i+=1
# Now we have k+1 possible eigenvalues and eigenvectors
# Return the ones specified by the keyword "which"
nreturned=iparam[4] # number of good eigenvalues returned
if nreturned==k: # we got exactly how many eigenvalues we wanted
d=d[:k]
z=z[:,:k]
else: # we got one extra eigenvalue (likely a cc pair, but which?)
# cut at approx precision for sorting
rd=np.round(d,decimals=_ndigits[typ])
if which in ['LR','SR']:
ind=np.argsort(rd.real)
elif which in ['LI','SI']:
# for LI,SI ARPACK returns largest,smallest abs(imaginary) why?
ind=np.argsort(abs(rd.imag))
else:
ind=np.argsort(abs(rd))
if which in ['LR','LM','LI']:
d=d[ind[-k:]]
z=z[:,ind[-k:]]
if which in ['SR','SM','SI']:
d=d[ind[:k]]
z=z[:,ind[:k]]
else:
# complex is so much simpler...
d,z,info =\
eigextract(rvec,howmny,sselect,sigmar,workev,
bmat,which,k,tol,resid,v,iparam,ipntr,
workd,workl,rwork,ierr)
if ierr != 0:
raise RuntimeError("Error info=%d in arpack"%info)
return None
if return_eigenvectors:
return d,z
return d
def eigen_symmetric(A, k=6, M=None, sigma=None, which='LM', v0=None,
ncv=None, maxiter=None, tol=0,
return_eigenvectors=True):
"""Find k eigenvalues and eigenvectors of the real symmetric
square matrix A.
Solves A * x[i] = w[i] * x[i], the standard eigenvalue problem for
w[i] eigenvalues with corresponding eigenvectors x[i].
Parameters
----------
A : matrix or array with real entries or object with matvec(x) method
An N x N real symmetric matrix or array or an object with matvec(x)
method to perform the matrix vector product A * x. The sparse
matrix formats in scipy.sparse are appropriate for A.
k : integer
The number of eigenvalues and eigenvectors desired
Returns
-------
w : array
Array of k eigenvalues
v : array
An array of k eigenvectors
The v[i] is the eigenvector corresponding to the eigenvector w[i]
Other Parameters
----------------
M : matrix or array
(Not implemented)
A symmetric positive-definite matrix for the generalized
eigenvalue problem A * x = w * M * x
sigma : real
(Not implemented)
Find eigenvalues near sigma. Shift spectrum by sigma.
v0 : array
Starting vector for iteration.
ncv : integer
The number of Lanczos vectors generated
ncv must be greater than k; it is recommended that ncv > 2*k
which : string
Which k eigenvectors and eigenvalues to find:
- 'LA' : Largest (algebraic) eigenvalues
- 'SA' : Smallest (algebraic) eigenvalues
- 'LM' : Largest (in magnitude) eigenvalues
- 'SM' : Smallest (in magnitude) eigenvalues
- 'BE' : Half (k/2) from each end of the spectrum
When k is odd, return one more (k/2+1) from the high end
maxiter : integer
Maximum number of Arnoldi update iterations allowed
tol : float
Relative accuracy for eigenvalues (stopping criterion)
return_eigenvectors : boolean
Return eigenvectors (True) in addition to eigenvalues
See Also
--------
eigen : eigenvalues and eigenvectors for a general (nonsymmetric) matrix A
Notes
-----
Examples
--------
"""
A = aslinearoperator(A)
if A.shape[0] != A.shape[1]:
raise ValueError('expected square matrix (shape=%s)' % (A.shape,))
n = A.shape[0]
if M is not None:
raise NotImplementedError("generalized eigenproblem not supported yet")
matvec = lambda x : A.matvec(x)
params = _SymmetricArpackParams(n, k, A.dtype.char, matvec, sigma,
ncv, v0, maxiter, which, tol)
while not params.converged:
params.iterate()
return params.extract(return_eigenvectors)
def make_system(A, M, x0, b):
"""Make a linear system Ax=b
Parameters
----------
A : LinearOperator
sparse or dense matrix (or any valid input to aslinearoperator)
M : {LinearOperator, Nones}
preconditioner
sparse or dense matrix (or any valid input to aslinearoperator)
x0 : {array_like, None}
initial guess to iterative method
b : array_like
right hand side
Returns
-------
(A, M, x, b, postprocess)
A : LinearOperator
matrix of the linear system
M : LinearOperator
preconditioner
x : rank 1 ndarray
initial guess
b : rank 1 ndarray
right hand side
postprocess : function
converts the solution vector to the appropriate
type and dimensions (e.g. (N,1) matrix)
"""
A_ = A
A = aslinearoperator(A)
if A.shape[0] != A.shape[1]:
raise ValueError('expected square matrix, but got shape=%s' % (A.shape,))
N = A.shape[0]
b = asanyarray(b)
if not (b.shape == (N,1) or b.shape == (N,)):
raise ValueError('A and b have incompatible dimensions')
if b.dtype.char not in 'fdFD':
b = b.astype('d') # upcast non-FP types to double
def postprocess(x):
if isinstance(b,matrix):
x = asmatrix(x)
return x.reshape(b.shape)
if hasattr(A,'dtype'):
xtype = A.dtype.char
else:
xtype = A.matvec(b).dtype.char
xtype = coerce(xtype, b.dtype.char)
b = asarray(b,dtype=xtype) # make b the same type as x
b = b.ravel()
if x0 is None:
x = zeros(N, dtype=xtype)
else:
x = array(x0, dtype=xtype)
if not (x.shape == (N,1) or x.shape == (N,)):
raise ValueError('A and x have incompatible dimensions')
x = x.ravel()
# process preconditioner
if M is None:
if hasattr(A_,'psolve'):
psolve = A_.psolve
else:
psolve = id
if hasattr(A_,'rpsolve'):
rpsolve = A_.rpsolve
else:
rpsolve = id
if psolve is id and rpsolve is id:
M = IdentityOperator(shape=A.shape, dtype=A.dtype)
else:
M = LinearOperator(A.shape, matvec=psolve, rmatvec=rpsolve,
dtype=A.dtype)
else:
M = aslinearoperator(M)
if A.shape != M.shape:
raise ValueError('matrix and preconditioner have different shapes')
return A, M, x, b, postprocess
def lsqr(A,
b,
damp=0.0,
atol=1e-8,
btol=1e-8,
conlim=1e8,
iter_lim=None,
show=False,
calc_var=False,
x0=None):
"""Find the least-squares solution to a large, sparse, linear system
of equations.
The function solves ``Ax = b`` or ``min ||b - Ax||^2`` or
``min ||Ax - b||^2 + d^2 ||x||^2``.
The matrix A may be square or rectangular (over-determined or
under-determined), and may have any rank.
::
1. Unsymmetric equations -- solve A*x = b
2. Linear least squares -- solve A*x = b
in the least-squares sense
3. Damped least squares -- solve ( A )*x = ( b )
( damp*I ) ( 0 )
in the least-squares sense
Parameters
----------
A : {sparse matrix, ndarray, LinearOperator}
Representation of an m-by-n matrix.
Alternatively, ``A`` can be a linear operator which can
produce ``Ax`` and ``A^T x`` using, e.g.,
``scipy.sparse.linalg.LinearOperator``.
b : array_like, shape (m,)
Right-hand side vector ``b``.
damp : float
Damping coefficient.
atol, btol : float, optional
Stopping tolerances. If both are 1.0e-9 (say), the final
residual norm should be accurate to about 9 digits. (The
final x will usually have fewer correct digits, depending on
cond(A) and the size of damp.)
conlim : float, optional
Another stopping tolerance. lsqr terminates if an estimate of
``cond(A)`` exceeds `conlim`. For compatible systems ``Ax =
b``, `conlim` could be as large as 1.0e+12 (say). For
least-squares problems, conlim should be less than 1.0e+8.
Maximum precision can be obtained by setting ``atol = btol =
conlim = zero``, but the number of iterations may then be
excessive.
iter_lim : int, optional
Explicit limitation on number of iterations (for safety).
show : bool, optional
Display an iteration log.
calc_var : bool, optional
Whether to estimate diagonals of ``(A'A + damp^2*I)^{-1}``.
x0 : array_like, shape (n,), optional
Initial guess of x, if None zeros are used.
.. versionadded:: 1.0.0
Returns
-------
x : ndarray of float
The final solution.
istop : int
Gives the reason for termination.
1 means x is an approximate solution to Ax = b.
2 means x approximately solves the least-squares problem.
itn : int
Iteration number upon termination.
r1norm : float
``norm(r)``, where ``r = b - Ax``.
r2norm : float
``sqrt( norm(r)^2 + damp^2 * norm(x)^2 )``. Equal to `r1norm` if
``damp == 0``.
anorm : float
Estimate of Frobenius norm of ``Abar = [[A]; [damp*I]]``.
acond : float
Estimate of ``cond(Abar)``.
arnorm : float
Estimate of ``norm(A'*r - damp^2*x)``.
xnorm : float
``norm(x)``
var : ndarray of float
If ``calc_var`` is True, estimates all diagonals of
``(A'A)^{-1}`` (if ``damp == 0``) or more generally ``(A'A +
damp^2*I)^{-1}``. This is well defined if A has full column
rank or ``damp > 0``. (Not sure what var means if ``rank(A)
< n`` and ``damp = 0.``)
Notes
-----
LSQR uses an iterative method to approximate the solution. The
number of iterations required to reach a certain accuracy depends
strongly on the scaling of the problem. Poor scaling of the rows
or columns of A should therefore be avoided where possible.
For example, in problem 1 the solution is unaltered by
row-scaling. If a row of A is very small or large compared to
the other rows of A, the corresponding row of ( A b ) should be
scaled up or down.
In problems 1 and 2, the solution x is easily recovered
following column-scaling. Unless better information is known,
the nonzero columns of A should be scaled so that they all have
the same Euclidean norm (e.g., 1.0).
In problem 3, there is no freedom to re-scale if damp is
nonzero. However, the value of damp should be assigned only
after attention has been paid to the scaling of A.
The parameter damp is intended to help regularize
ill-conditioned systems, by preventing the true solution from
being very large. Another aid to regularization is provided by
the parameter acond, which may be used to terminate iterations
before the computed solution becomes very large.
If some initial estimate ``x0`` is known and if ``damp == 0``,
one could proceed as follows:
1. Compute a residual vector ``r0 = b - A*x0``.
2. Use LSQR to solve the system ``A*dx = r0``.
3. Add the correction dx to obtain a final solution ``x = x0 + dx``.
This requires that ``x0`` be available before and after the call
to LSQR. To judge the benefits, suppose LSQR takes k1 iterations
to solve A*x = b and k2 iterations to solve A*dx = r0.
If x0 is "good", norm(r0) will be smaller than norm(b).
If the same stopping tolerances atol and btol are used for each
system, k1 and k2 will be similar, but the final solution x0 + dx
should be more accurate. The only way to reduce the total work
is to use a larger stopping tolerance for the second system.
If some value btol is suitable for A*x = b, the larger value
btol*norm(b)/norm(r0) should be suitable for A*dx = r0.
Preconditioning is another way to reduce the number of iterations.
If it is possible to solve a related system ``M*x = b``
efficiently, where M approximates A in some helpful way (e.g. M -
A has low rank or its elements are small relative to those of A),
LSQR may converge more rapidly on the system ``A*M(inverse)*z =
b``, after which x can be recovered by solving M*x = z.
If A is symmetric, LSQR should not be used!
Alternatives are the symmetric conjugate-gradient method (cg)
and/or SYMMLQ. SYMMLQ is an implementation of symmetric cg that
applies to any symmetric A and will converge more rapidly than
LSQR. If A is positive definite, there are other implementations
of symmetric cg that require slightly less work per iteration than
SYMMLQ (but will take the same number of iterations).
References
----------
.. [1] C. C. Paige and M. A. Saunders (1982a).
"LSQR: An algorithm for sparse linear equations and
sparse least squares", ACM TOMS 8(1), 43-71.
.. [2] C. C. Paige and M. A. Saunders (1982b).
"Algorithm 583. LSQR: Sparse linear equations and least
squares problems", ACM TOMS 8(2), 195-209.
.. [3] M. A. Saunders (1995). "Solution of sparse rectangular
systems using LSQR and CRAIG", BIT 35, 588-604.
Examples
--------
>>> from scipy.sparse import csc_matrix
>>> from scipy.sparse.linalg import lsqr
>>> A = csc_matrix([[1., 0.], [1., 1.], [0., 1.]], dtype=float)
The first example has the trivial solution `[0, 0]`
>>> b = np.array([0., 0., 0.], dtype=float)
>>> x, istop, itn, normr = lsqr(A, b)[:4]
The exact solution is x = 0
>>> istop
0
>>> x
array([ 0., 0.])
The stopping code `istop=0` returned indicates that a vector of zeros was
found as a solution. The returned solution `x` indeed contains `[0., 0.]`.
The next example has a non-trivial solution:
>>> b = np.array([1., 0., -1.], dtype=float)
>>> x, istop, itn, r1norm = lsqr(A, b)[:4]
>>> istop
1
>>> x
array([ 1., -1.])
>>> itn
1
>>> r1norm
4.440892098500627e-16
As indicated by `istop=1`, `lsqr` found a solution obeying the tolerance
limits. The given solution `[1., -1.]` obviously solves the equation. The
remaining return values include information about the number of iterations
(`itn=1`) and the remaining difference of left and right side of the solved
equation.
The final example demonstrates the behavior in the case where there is no
solution for the equation:
>>> b = np.array([1., 0.01, -1.], dtype=float)
>>> x, istop, itn, r1norm = lsqr(A, b)[:4]
>>> istop
2
>>> x
array([ 1.00333333, -0.99666667])
>>> A.dot(x)-b
array([ 0.00333333, -0.00333333, 0.00333333])
>>> r1norm
0.005773502691896255
`istop` indicates that the system is inconsistent and thus `x` is rather an
approximate solution to the corresponding least-squares problem. `r1norm`
contains the norm of the minimal residual that was found.
"""
A = aslinearoperator(A)
b = np.atleast_1d(b)
if b.ndim > 1:
b = b.squeeze()
m, n = A.shape
if iter_lim is None:
iter_lim = 2 * n
var = np.zeros(n)
msg = ('The exact solution is x = 0 ',
'Ax - b is small enough, given atol, btol ',
'The least-squares solution is good enough, given atol ',
'The estimate of cond(Abar) has exceeded conlim ',
'Ax - b is small enough for this machine ',
'The least-squares solution is good enough for this machine',
'Cond(Abar) seems to be too large for this machine ',
'The iteration limit has been reached ')
if show:
print(' ')
print('LSQR Least-squares solution of Ax = b')
str1 = 'The matrix A has %8g rows and %8g cols' % (m, n)
str2 = 'damp = %20.14e calc_var = %8g' % (damp, calc_var)
str3 = 'atol = %8.2e conlim = %8.2e' % (atol, conlim)
str4 = 'btol = %8.2e iter_lim = %8g' % (btol, iter_lim)
print(str1)
print(str2)
print(str3)
print(str4)
itn = 0
istop = 0
ctol = 0
if conlim > 0:
ctol = 1 / conlim
anorm = 0
acond = 0
dampsq = damp**2
ddnorm = 0
res2 = 0
xnorm = 0
xxnorm = 0
z = 0
cs2 = -1
sn2 = 0
"""
Set up the first vectors u and v for the bidiagonalization.
These satisfy beta*u = b - A*x, alfa*v = A'*u.
"""
u = b
bnorm = np.linalg.norm(b)
if x0 is None:
x = np.zeros(n)
beta = bnorm.copy()
else:
x = np.asarray(x0)
u = u - A.matvec(x)
beta = np.linalg.norm(u)
if beta > 0:
u = (1 / beta) * u
v = A.rmatvec(u)
alfa = np.linalg.norm(v)
else:
v = x.copy()
alfa = 0
if alfa > 0:
v = (1 / alfa) * v
w = v.copy()
rhobar = alfa
phibar = beta
rnorm = beta
r1norm = rnorm
r2norm = rnorm
# Reverse the order here from the original matlab code because
# there was an error on return when arnorm==0
arnorm = alfa * beta
if arnorm == 0:
print(msg[0])
return x, istop, itn, r1norm, r2norm, anorm, acond, arnorm, xnorm, var
head1 = ' Itn x[0] r1norm r2norm '
head2 = ' Compatible LS Norm A Cond A'
if show:
print(' ')
print(head1, head2)
test1 = 1
test2 = alfa / beta
str1 = '%6g %12.5e' % (itn, x[0])
str2 = ' %10.3e %10.3e' % (r1norm, r2norm)
str3 = ' %8.1e %8.1e' % (test1, test2)
print(str1, str2, str3)
# Main iteration loop.
while itn < iter_lim:
itn = itn + 1
"""
% Perform the next step of the bidiagonalization to obtain the
% next beta, u, alfa, v. These satisfy the relations
% beta*u = a*v - alfa*u,
% alfa*v = A'*u - beta*v.
"""
u = A.matvec(v) - alfa * u
beta = np.linalg.norm(u)
if beta > 0:
u = (1 / beta) * u
anorm = sqrt(anorm**2 + alfa**2 + beta**2 + damp**2)
v = A.rmatvec(u) - beta * v
alfa = np.linalg.norm(v)
if alfa > 0:
v = (1 / alfa) * v
# Use a plane rotation to eliminate the damping parameter.
# This alters the diagonal (rhobar) of the lower-bidiagonal matrix.
rhobar1 = sqrt(rhobar**2 + damp**2)
cs1 = rhobar / rhobar1
sn1 = damp / rhobar1
psi = sn1 * phibar
phibar = cs1 * phibar
# Use a plane rotation to eliminate the subdiagonal element (beta)
# of the lower-bidiagonal matrix, giving an upper-bidiagonal matrix.
cs, sn, rho = _sym_ortho(rhobar1, beta)
theta = sn * alfa
rhobar = -cs * alfa
phi = cs * phibar
phibar = sn * phibar
tau = sn * phi
# Update x and w.
t1 = phi / rho
t2 = -theta / rho
dk = (1 / rho) * w
x = x + t1 * w
w = v + t2 * w
ddnorm = ddnorm + np.linalg.norm(dk)**2
if calc_var:
var = var + dk**2
# Use a plane rotation on the right to eliminate the
# super-diagonal element (theta) of the upper-bidiagonal matrix.
# Then use the result to estimate norm(x).
delta = sn2 * rho
gambar = -cs2 * rho
rhs = phi - delta * z
zbar = rhs / gambar
xnorm = sqrt(xxnorm + zbar**2)
gamma = sqrt(gambar**2 + theta**2)
cs2 = gambar / gamma
sn2 = theta / gamma
z = rhs / gamma
xxnorm = xxnorm + z**2
# Test for convergence.
# First, estimate the condition of the matrix Abar,
# and the norms of rbar and Abar'rbar.
acond = anorm * sqrt(ddnorm)
res1 = phibar**2
res2 = res2 + psi**2
rnorm = sqrt(res1 + res2)
arnorm = alfa * abs(tau)
# Distinguish between
# r1norm = ||b - Ax|| and
# r2norm = rnorm in current code
# = sqrt(r1norm^2 + damp^2*||x||^2).
# Estimate r1norm from
# r1norm = sqrt(r2norm^2 - damp^2*||x||^2).
# Although there is cancellation, it might be accurate enough.
r1sq = rnorm**2 - dampsq * xxnorm
r1norm = sqrt(abs(r1sq))
if r1sq < 0:
r1norm = -r1norm
r2norm = rnorm
# Now use these norms to estimate certain other quantities,
# some of which will be small near a solution.
test1 = rnorm / bnorm
test2 = arnorm / (anorm * rnorm + eps)
test3 = 1 / (acond + eps)
t1 = test1 / (1 + anorm * xnorm / bnorm)
rtol = btol + atol * anorm * xnorm / bnorm
# The following tests guard against extremely small values of
# atol, btol or ctol. (The user may have set any or all of
# the parameters atol, btol, conlim to 0.)
# The effect is equivalent to the normal tests using
# atol = eps, btol = eps, conlim = 1/eps.
if itn >= iter_lim:
istop = 7
if 1 + test3 <= 1:
istop = 6
if 1 + test2 <= 1:
istop = 5
if 1 + t1 <= 1:
istop = 4
# Allow for tolerances set by the user.
if test3 <= ctol:
istop = 3
if test2 <= atol:
istop = 2
if test1 <= rtol:
istop = 1
# See if it is time to print something.
prnt = False
if n <= 40:
prnt = True
if itn <= 10:
prnt = True
if itn >= iter_lim - 10:
prnt = True
# if itn%10 == 0: prnt = True
if test3 <= 2 * ctol:
prnt = True
if test2 <= 10 * atol:
prnt = True
if test1 <= 10 * rtol:
prnt = True
if istop != 0:
prnt = True
if prnt:
if show:
str1 = '%6g %12.5e' % (itn, x[0])
str2 = ' %10.3e %10.3e' % (r1norm, r2norm)
str3 = ' %8.1e %8.1e' % (test1, test2)
str4 = ' %8.1e %8.1e' % (anorm, acond)
print(str1, str2, str3, str4)
if istop != 0:
break
# End of iteration loop.
# Print the stopping condition.
if show:
print(' ')
print('LSQR finished')
print(msg[istop])
print(' ')
str1 = 'istop =%8g r1norm =%8.1e' % (istop, r1norm)
str2 = 'anorm =%8.1e arnorm =%8.1e' % (anorm, arnorm)
str3 = 'itn =%8g r2norm =%8.1e' % (itn, r2norm)
str4 = 'acond =%8.1e xnorm =%8.1e' % (acond, xnorm)
print(str1 + ' ' + str2)
print(str3 + ' ' + str4)
print(' ')
return x, istop, itn, r1norm, r2norm, anorm, acond, arnorm, xnorm, var
def assertCompatibleSystem(self, A, xtrue):
Afun = aslinearoperator(A)
b = Afun.matvec(xtrue)
x = lsmr(A, b)[0]
assert_almost_equal(norm(x - xtrue), 0, decimal=5)
def cgne(A,
b,
x0=None,
tol=1e-5,
maxiter=None,
xtype=None,
M=None,
callback=None,
residuals=None):
'''Conjugate Gradient, Normal Error algorithm
Applies CG to the normal equations, A.H A x = b. Left preconditioning
is supported. Note that unless A is well-conditioned, the use of
CGNE is inadvisable
Parameters
----------
A : {array, matrix, sparse matrix, LinearOperator}
n x n, linear system to solve
b : {array, matrix}
right hand side, shape is (n,) or (n,1)
x0 : {array, matrix}
initial guess, default is a vector of zeros
tol : float
relative convergence tolerance, i.e. tol is scaled by ||r_0||_2
maxiter : int
maximum number of allowed iterations
xtype : type
dtype for the solution, default is automatic type detection
M : {array, matrix, sparse matrix, LinearOperator}
n x n, inverted preconditioner, i.e. solve M A A.H x = M b.
callback : function
User-supplied function is called after each iteration as
callback(xk), where xk is the current solution vector
residuals : list
residuals has the residual norm history,
including the initial residual, appended to it
Returns
-------
(xNew, info)
xNew : an updated guess to the solution of Ax = b
info : halting status of cgne
== =======================================
0 successful exit
>0 convergence to tolerance not achieved,
return iteration count instead.
<0 numerical breakdown, or illegal input
== =======================================
Notes
-----
- The LinearOperator class is in scipy.sparse.linalg.interface.
Use this class if you prefer to define A or M as a mat-vec routine
as opposed to explicitly constructing the matrix. A.psolve(..) is
still supported as a legacy.
Examples
--------
>>> from pyamg.krylov.cgne import cgne
>>> from pyamg.util.linalg import norm
>>> import numpy as np
>>> from pyamg.gallery import poisson
>>> A = poisson((10,10))
>>> b = np.ones((A.shape[0],))
>>> (x,flag) = cgne(A,b, maxiter=2, tol=1e-8)
>>> print norm(b - A*x)
46.1547104367
References
----------
.. [1] Yousef Saad, "Iterative Methods for Sparse Linear Systems,
Second Edition", SIAM, pp. 276-7, 2003
http://www-users.cs.umn.edu/~saad/books.html
'''
# Store the conjugate transpose explicitly as it will be used much later on
if isspmatrix(A):
AH = A.H
else:
# TODO avoid doing this since A may be a different sparse type
AH = aslinearoperator(np.asmatrix(A).H)
# Convert inputs to linear system, with error checking
A, M, x, b, postprocess = make_system(A, M, x0, b, xtype)
dimen = A.shape[0]
# Ensure that warnings are always reissued from this function
import warnings
warnings.filterwarnings('always', module='pyamg\.krylov\._cgne')
# Choose type
if not hasattr(A, 'dtype'):
Atype = upcast(x.dtype, b.dtype)
else:
Atype = A.dtype
if not hasattr(M, 'dtype'):
Mtype = upcast(x.dtype, b.dtype)
else:
Mtype = M.dtype
xtype = upcast(Atype, x.dtype, b.dtype, Mtype)
# Should norm(r) be kept
if residuals == []:
keep_r = True
else:
keep_r = False
# How often should r be recomputed
recompute_r = 8
# Check iteration numbers. CGNE suffers from loss of orthogonality quite
# easily, so we arbitrarily let the method go up to 130% over the
# theoretically necessary limit of maxiter=dimen
if maxiter is None:
maxiter = int(np.ceil(1.3 * dimen)) + 2
elif maxiter < 1:
raise ValueError('Number of iterations must be positive')
elif maxiter > (1.3 * dimen):
warn('maximum allowed inner iterations (maxiter) are the 130% times \
the number of dofs')
maxiter = int(np.ceil(1.3 * dimen)) + 2
# Prep for method
r = b - A * x
normr = norm(r)
if keep_r:
residuals.append(normr)
# Check initial guess ( scaling by b, if b != 0,
# must account for case when norm(b) is very small)
normb = norm(b)
if normb == 0.0:
normb = 1.0
if normr < tol * normb:
if callback is not None:
callback(x)
return (postprocess(x), 0)
# Scale tol by ||r_0||_2
if normr != 0.0:
tol = tol * normr
# Begin CGNE
# Apply preconditioner and calculate initial search direction
z = M * r
p = AH * z
old_zr = np.inner(z.conjugate(), r)
for iter in range(maxiter):
# alpha = (z_j, r_j) / (p_j, p_j)
alpha = old_zr / np.inner(p.conjugate(), p)
# x_{j+1} = x_j + alpha*p_j
x += alpha * p
# r_{j+1} = r_j - alpha*w_j, where w_j = A*p_j
if np.mod(iter, recompute_r) and iter > 0:
r -= alpha * (A * p)
else:
r = b - A * x
# z_{j+1} = M*r_{j+1}
z = M * r
# beta = (z_{j+1}, r_{j+1}) / (z_j, r_j)
new_zr = np.inner(z.conjugate(), r)
beta = new_zr / old_zr
old_zr = new_zr
# p_{j+1} = A.H*z_{j+1} + beta*p_j
p *= beta
p += AH * z
# Allow user access to residual
if callback is not None:
callback(x)
# test for convergence
normr = norm(r)
if keep_r:
residuals.append(normr)
if normr < tol:
return (postprocess(x), 0)
# end loop
return (postprocess(x), iter + 1)
def lsmr(A,
b,
damp=0.0,
atol=1e-6,
btol=1e-6,
conlim=1e8,
maxiter=None,
show=False):
"""Iterative solver for least-squares problems.
lsmr solves the system of linear equations ``Ax = b``. If the system
is inconsistent, it solves the least-squares problem ``min ||b - Ax||_2``.
A is a rectangular matrix of dimension m-by-n, where all cases are
allowed: m = n, m > n, or m < n. B is a vector of length m.
The matrix A may be dense or sparse (usually sparse).
.. versionadded:: 0.11.0
Parameters
----------
A : {matrix, sparse matrix, ndarray, LinearOperator}
Matrix A in the linear system.
b : (m,) ndarray
Vector b in the linear system.
damp : float
Damping factor for regularized least-squares. `lsmr` solves
the regularized least-squares problem::
min ||(b) - ( A )x||
||(0) (damp*I) ||_2
where damp is a scalar. If damp is None or 0, the system
is solved without regularization.
atol, btol : float
Stopping tolerances. `lsmr` continues iterations until a
certain backward error estimate is smaller than some quantity
depending on atol and btol. Let ``r = b - Ax`` be the
residual vector for the current approximate solution ``x``.
If ``Ax = b`` seems to be consistent, ``lsmr`` terminates
when ``norm(r) <= atol * norm(A) * norm(x) + btol * norm(b)``.
Otherwise, lsmr terminates when ``norm(A^{T} r) <=
atol * norm(A) * norm(r)``. If both tolerances are 1.0e-6 (say),
the final ``norm(r)`` should be accurate to about 6
digits. (The final x will usually have fewer correct digits,
depending on ``cond(A)`` and the size of LAMBDA.) If `atol`
or `btol` is None, a default value of 1.0e-6 will be used.
Ideally, they should be estimates of the relative error in the
entries of A and B respectively. For example, if the entries
of `A` have 7 correct digits, set atol = 1e-7. This prevents
the algorithm from doing unnecessary work beyond the
uncertainty of the input data.
conlim : float
`lsmr` terminates if an estimate of ``cond(A)`` exceeds
`conlim`. For compatible systems ``Ax = b``, conlim could be
as large as 1.0e+12 (say). For least-squares problems,
`conlim` should be less than 1.0e+8. If `conlim` is None, the
default value is 1e+8. Maximum precision can be obtained by
setting ``atol = btol = conlim = 0``, but the number of
iterations may then be excessive.
maxiter : int
`lsmr` terminates if the number of iterations reaches
`maxiter`. The default is ``maxiter = min(m, n)``. For
ill-conditioned systems, a larger value of `maxiter` may be
needed.
show : bool
Print iterations logs if ``show=True``.
Returns
-------
x : ndarray of float
Least-square solution returned.
istop : int
istop gives the reason for stopping::
istop = 0 means x=0 is a solution.
= 1 means x is an approximate solution to A*x = B,
according to atol and btol.
= 2 means x approximately solves the least-squares problem
according to atol.
= 3 means COND(A) seems to be greater than CONLIM.
= 4 is the same as 1 with atol = btol = eps (machine
precision)
= 5 is the same as 2 with atol = eps.
= 6 is the same as 3 with CONLIM = 1/eps.
= 7 means ITN reached maxiter before the other stopping
conditions were satisfied.
itn : int
Number of iterations used.
normr : float
``norm(b-Ax)``
normar : float
``norm(A^T (b - Ax))``
norma : float
``norm(A)``
conda : float
Condition number of A.
normx : float
``norm(x)``
References
----------
.. [1] D. C.-L. Fong and M. A. Saunders,
"LSMR: An iterative algorithm for sparse least-squares problems",
SIAM J. Sci. Comput., vol. 33, pp. 2950-2971, 2011.
http://arxiv.org/abs/1006.0758
.. [2] LSMR Software, http://www.stanford.edu/~clfong/lsmr.html
"""
A = aslinearoperator(A)
b = b.squeeze()
msg = ('The exact solution is x = 0 ',
'Ax - b is small enough, given atol, btol ',
'The least-squares solution is good enough, given atol ',
'The estimate of cond(Abar) has exceeded conlim ',
'Ax - b is small enough for this machine ',
'The least-squares solution is good enough for this machine',
'Cond(Abar) seems to be too large for this machine ',
'The iteration limit has been reached ')
hdg1 = ' itn x(1) norm r norm A' 'r'
hdg2 = ' compatible LS norm A cond A'
pfreq = 20 # print frequency (for repeating the heading)
pcount = 0 # print counter
m, n = A.shape
# stores the num of singular values
minDim = min([m, n])
if maxiter is None:
maxiter = minDim
if show:
print(' ')
print('LSMR Least-squares solution of Ax = b\n')
print('The matrix A has %8g rows and %8g cols' % (m, n))
print('damp = %20.14e\n' % (damp))
print('atol = %8.2e conlim = %8.2e\n' % (atol, conlim))
print('btol = %8.2e maxiter = %8g\n' % (btol, maxiter))
u = b
beta = norm(u)
v = zeros(n)
alpha = 0
if beta > 0:
u = (1 / beta) * u
v = A.rmatvec(u)
alpha = norm(v)
if alpha > 0:
v = (1 / alpha) * v
# Initialize variables for 1st iteration.
itn = 0
zetabar = alpha * beta
alphabar = alpha
rho = 1
rhobar = 1
cbar = 1
sbar = 0
h = v.copy()
hbar = zeros(n)
x = zeros(n)
# Initialize variables for estimation of ||r||.
betadd = beta
betad = 0
rhodold = 1
tautildeold = 0
thetatilde = 0
zeta = 0
d = 0
# Initialize variables for estimation of ||A|| and cond(A)
normA2 = alpha * alpha
maxrbar = 0
minrbar = 1e+100
normA = sqrt(normA2)
condA = 1
normx = 0
# Items for use in stopping rules.
normb = beta
istop = 0
ctol = 0
if conlim > 0:
ctol = 1 / conlim
normr = beta
# Reverse the order here from the original matlab code because
# there was an error on return when arnorm==0
normar = alpha * beta
if normar == 0:
if show:
print(msg[0])
return x, istop, itn, normr, normar, normA, condA, normx
if show:
print(' ')
print(hdg1, hdg2)
test1 = 1
test2 = alpha / beta
str1 = '%6g %12.5e' % (itn, x[0])
str2 = ' %10.3e %10.3e' % (normr, normar)
str3 = ' %8.1e %8.1e' % (test1, test2)
print(''.join([str1, str2, str3]))
# Main iteration loop.
while itn < maxiter:
itn = itn + 1
# Perform the next step of the bidiagonalization to obtain the
# next beta, u, alpha, v. These satisfy the relations
# beta*u = a*v - alpha*u,
# alpha*v = A'*u - beta*v.
u = A.matvec(v) - alpha * u
beta = norm(u)
if beta > 0:
u = (1 / beta) * u
v = A.rmatvec(u) - beta * v
alpha = norm(v)
if alpha > 0:
v = (1 / alpha) * v
# At this point, beta = beta_{k+1}, alpha = alpha_{k+1}.
# Construct rotation Qhat_{k,2k+1}.
chat, shat, alphahat = _sym_ortho(alphabar, damp)
# Use a plane rotation (Q_i) to turn B_i to R_i
rhoold = rho
c, s, rho = _sym_ortho(alphahat, beta)
thetanew = s * alpha
alphabar = c * alpha
# Use a plane rotation (Qbar_i) to turn R_i^T to R_i^bar
rhobarold = rhobar
zetaold = zeta
thetabar = sbar * rho
rhotemp = cbar * rho
cbar, sbar, rhobar = _sym_ortho(cbar * rho, thetanew)
zeta = cbar * zetabar
zetabar = -sbar * zetabar
# Update h, h_hat, x.
hbar = h - (thetabar * rho / (rhoold * rhobarold)) * hbar
x = x + (zeta / (rho * rhobar)) * hbar
h = v - (thetanew / rho) * h
# Estimate of ||r||.
# Apply rotation Qhat_{k,2k+1}.
betaacute = chat * betadd
betacheck = -shat * betadd
# Apply rotation Q_{k,k+1}.
betahat = c * betaacute
betadd = -s * betaacute
# Apply rotation Qtilde_{k-1}.
# betad = betad_{k-1} here.
thetatildeold = thetatilde
ctildeold, stildeold, rhotildeold = _sym_ortho(rhodold, thetabar)
thetatilde = stildeold * rhobar
rhodold = ctildeold * rhobar
betad = -stildeold * betad + ctildeold * betahat
# betad = betad_k here.
# rhodold = rhod_k here.
tautildeold = (zetaold - thetatildeold * tautildeold) / rhotildeold
taud = (zeta - thetatilde * tautildeold) / rhodold
d = d + betacheck * betacheck
normr = sqrt(d + (betad - taud)**2 + betadd * betadd)
# Estimate ||A||.
normA2 = normA2 + beta * beta
normA = sqrt(normA2)
normA2 = normA2 + alpha * alpha
# Estimate cond(A).
maxrbar = max(maxrbar, rhobarold)
if itn > 1:
minrbar = min(minrbar, rhobarold)
condA = max(maxrbar, rhotemp) / min(minrbar, rhotemp)
# Test for convergence.
# Compute norms for convergence testing.
normar = abs(zetabar)
normx = norm(x)
# Now use these norms to estimate certain other quantities,
# some of which will be small near a solution.
test1 = normr / normb
if (normA * normr) != 0:
test2 = normar / (normA * normr)
else:
test2 = infty
test3 = 1 / condA
t1 = test1 / (1 + normA * normx / normb)
rtol = btol + atol * normA * normx / normb
# The following tests guard against extremely small values of
# atol, btol or ctol. (The user may have set any or all of
# the parameters atol, btol, conlim to 0.)
# The effect is equivalent to the normAl tests using
# atol = eps, btol = eps, conlim = 1/eps.
if itn >= maxiter:
istop = 7
if 1 + test3 <= 1:
istop = 6
if 1 + test2 <= 1:
istop = 5
if 1 + t1 <= 1:
istop = 4
# Allow for tolerances set by the user.
if test3 <= ctol:
istop = 3
if test2 <= atol:
istop = 2
if test1 <= rtol:
istop = 1
# See if it is time to print something.
if show:
if (n <= 40) or (itn <= 10) or (itn >= maxiter - 10) or \
(itn % 10 == 0) or (test3 <= 1.1 * ctol) or \
(test2 <= 1.1 * atol) or (test1 <= 1.1 * rtol) or \
(istop != 0):
if pcount >= pfreq:
pcount = 0
print(' ')
print(hdg1, hdg2)
pcount = pcount + 1
str1 = '%6g %12.5e' % (itn, x[0])
str2 = ' %10.3e %10.3e' % (normr, normar)
str3 = ' %8.1e %8.1e' % (test1, test2)
str4 = ' %8.1e %8.1e' % (normA, condA)
print(''.join([str1, str2, str3, str4]))
if istop > 0:
break
# Print the stopping condition.
if show:
print(' ')
print('LSMR finished')
print(msg[istop])
print('istop =%8g normr =%8.1e' % (istop, normr))
print(' normA =%8.1e normAr =%8.1e' % (normA, normar))
print('itn =%8g condA =%8.1e' % (itn, condA))
print(' normx =%8.1e' % (normx))
print(str1, str2)
print(str3, str4)
return x, istop, itn, normr, normar, normA, condA, normx
def mlem(y,
A,
no_iter,
verbose=False,
ret_iter_x=0,
ret_iter_y=0,
ret_norm_r=False,
ret_objective=False,
AT_ones=None,
x0=None,
inverse_thres=0.0):
'''
Maximizes the log-likelihood for a Poisson Random Varible. y is the
observed poisson random variable, modeled by A * x. Maximizes
f(x) = 1^T Ax - y^T ln(Ax)
Taken from Jingyu Cui's Thesis, "Fast and Accurate PET Image Reconstruction
on Parallel Architectures," 2013.
Parameters
----------
y : (m,) array-like
Observed Poisson variable
A : (m,n) matrix, sparse matrix, or LinearOperator
System Model
no_iter : int scalar
Number of update iterations to perform.
verbose : boolean (Default = False), optional
If the relative residual norm and objective should be printed out
each iteration.
ret_iter_x : int scalar, optional
Return the inter-iteration history of x every from ret_iter_x
iterations. If zero, the inter-iteration history of x is not returned
ret_iter_y : int scalar, optional
Return the inter-iteration history of y_bar (the model) every from
ret_iter_y iterations. If zero, the inter-iteration history of y_bar
is not returned.
ret_norm_r : boolean (Default = False), optional
Return the norm of the relative residual from iteration.
ret_objective : boolean (Default = False), optional
Return the objective function value from iteration.
AT_ones : (n,) array-like, optional
The result of A^T 1 can be provided to avoid the computation. AT_ones
is used to normalize the error backpropogation step.
x0 : (n,) array-like, optional
Override the default model initialization. Default is the number of
counts in y, divided by n for each x0.
inverse_thres : float scalar
Zeros out small inverse values for the error propogation. The inverse
of errors are backprojected with A^T. Small model values can cause the
value to explode unnecessarily.
Returns
-------
x : (n,) ndarray
The weights resulting from the algorithm.
x_history : (variable, n) ndarray, optional
The x from each iteration specified by ret_iter_x, plus the final value.
y_history : (variable, m) ndarray, optional
The y_bar from each iteration specified by ret_iter_y, plus the final
model value.
norm_r_history : (no_iter,) ndarray, optional
The norm of the relative residual from iteration
objective_history : (no_iter,) ndarray, optional
The objective function value from iteration
'''
A = aslinearoperator(A)
y = np.asarray(y, dtype=A.dtype).squeeze()
if AT_ones is None:
AT_ones = A.rmatvec(np.ones(A.shape[0]))
else:
AT_ones = np.asarray(AT_ones, dtype=A.dtype).squeeze()
if AT_ones.shape != (A.shape[1], ):
raise ValueError('AT_ones is not shaped (%d,)' % A.shape[1])
if x0 is None:
# Initialize it to uniform weights where the total counts would match
x = np.ones(A.shape[1], dtype=A.dtype) * (y.sum() / A.shape[1])
else:
x = np.asarray(x0, dtype=A.dtype).squeeze()
norm_y = np.linalg.norm(y)
# Save every history_idx iterations, and the last one
save_model_idx = np.zeros(no_iter + 1, dtype=bool)
if ret_iter_y > 0:
save_model_idx[(np.arange(no_iter + 1) % ret_iter_y) == 0] = True
save_model_idx[-1] = True
save_x_idx = np.zeros(no_iter + 1, dtype=bool)
if ret_iter_x > 0:
save_x_idx[(np.arange(no_iter + 1) % ret_iter_x) == 0] = True
save_x_idx[-1] = True
history_size_model = np.sum(save_model_idx)
history_size_x = np.sum(save_x_idx)
x_history = np.zeros((history_size_x, A.shape[1]))
model_history = np.zeros((history_size_model, A.shape[0]))
# These vectors are really small in comparison, so save them every
# iteration, and worry about returning them later.
norm_r_history = np.zeros((no_iter + 1, ))
objective_history = np.zeros((no_iter + 1, ))
history_count_x = 0
history_count_model = 0
for iter_no in xrange(no_iter + 1):
model = A.matvec(x)
norm_r = np.linalg.norm(model - y) / norm_y
objective = model[model > 0].astype(np.float128).sum() - \
(y[model > 0] * np.log(model[model > 0])
).astype(np.float128).sum()
if verbose:
print('{0:02d}: rel_norm = {1}, objective = {2}'.format(
iter_no, norm_r, objective))
if save_x_idx[iter_no]:
x_history[history_count_x, :] = x.copy()
history_count_x += 1
if save_model_idx[iter_no]:
model_history[history_count_model, :] = model.copy()
history_count_model += 1
norm_r_history[iter_no] = norm_r
objective_history[iter_no] = objective
# We loop for no_iter + 1 so that we can calculate the model error
# for the final iteration.
if iter_no == no_iter:
continue
error = np.zeros(A.shape[0], dtype=A.dtype)
error[model > 0] = y[model > 0] / model[model > 0]
error[model <= 0] = 0
if inverse_thres > 0:
error[error > 1.0 / inverse_thres] = 0
error_bp = A.rmatvec(error)
update = np.zeros(A.shape[1], dtype=A.dtype)
update[AT_ones > 0] = error_bp[AT_ones > 0] / AT_ones[AT_ones > 0]
update[AT_ones <= 0] = 0
update[update < 0] = 0
x *= update
ret = [
x,
]
if ret_iter_x > 0:
ret.append(x_history)
if ret_iter_y > 0:
ret.append(model_history)
if ret_norm_r:
ret.append(norm_r_history)
if ret_objective:
ret.append(objective_history)
if len(ret) == 1:
ret = ret[0]
return ret
def svds(A,
k=6,
ncv=None,
tol=0,
which='LM',
v0=None,
maxiter=None,
return_singular_vectors=True,
precAHA=None,
precAAH=None,
precAug=None,
u0=None,
locku0=None,
lockv0=None,
return_stats=False,
maxBlockSize=0):
"""
Compute k singular values and vectors for a sparse matrix.
Parameters
----------
A : {sparse matrix, LinearOperator}
Array to compute the SVD on, of shape (M, N)
k : int, optional
Number of singular values and vectors to compute.
Must be 1 <= k < min(A.shape).
ncv : int, optional
The maximum size of the basis
tol : float, optional
Tolerance for singular values. Zero (default) means machine precision.
which : str ['LM' | 'SM'] or number, optional
Which `k` singular values to find:
- 'LM' : largest singular values
- 'SM' : smallest singular values
- number : closest singular values to (referred as sigma later)
u0 : ndarray, optional
Left starting vectors for the iterations.
Should be approximate left singular vectors. If only u0 or v0 is
provided, the other is computed.
v0 : ndarray, optional
Right starting vectors for the iterations.
maxiter : int, optional
Maximum number of iterations.
precAHA : {N x N matrix, array, sparse matrix, LinearOperator}, optional
Approximate inverse of (A.H*A - sigma**2*I). If provided and M>N, it
usually accelerates the convergence.
precAAH : {M x M matrix, array, sparse matrix, LinearOperator}, optional
Approximate inverse of (A*A.H - sigma**2*I). If provided and M<N, it
usually accelerates the convergence.
precAug : {(M+N) x (M+N) matrix, array, sparse matrix, LinearOperator}, optional
Approximate inverse of ([zeros() A.H; zeros() A] - sigma*I). It usually
accelerates the convergence if tol<dtype.eps**.5.
locku0 : ndarray, optional
Left orthogonal vector constrain.
Seek singular triplets orthogonal to locku0 and lockv0. The provided vectors
*should* be orthonormal. If only locku0 or lockv0 is provided, the other
is computed. Useful to not converge some already computed solutions.
lockv0 : ndarray, optional
Right orthogonal vector constrain. See locku0.
maxBlockSize : int, optional
Maximum number of vectors added at every iteration.
report_stats : bool, optional
If True, it is also returned extra information from PRIMME.
Returns
-------
u : ndarray, shape=(M, k), optional
Unitary matrix having left singular vectors as columns.
Returned if `return_singular_vectors` is True.
s : ndarray, shape=(k,)
The singular values.
vt : ndarray, shape=(k, N), optional
Unitary matrix having right singular vectors as rows.
Returned if `return_singular_vectors` is True.
stats : dict, optional (if return_stats)
Extra information reported by PRIMME:
- "numOuterIterations": number of outer iterations
- "numRestarts": number of restarts
- "numMatvecs": number of A*v
- "numPreconds": number of OPinv*v
- "elapsedTime": time that took
Returned if `return_stats` is True.
See Also
--------
Primme.eigsh : eigenvalue decomposition for a sparse symmetrix/complex Hermitian matrix A
scipy.sparse.linalg.eigs : eigenvalues and eigenvectors for a general (nonsymmetric) matrix A
Examples
--------
>>> import Primme, scipy.sparse
>>> A = scipy.sparse.spdiags(range(10), [0], 100, 10) # sparse diag. rect. matrix
>>> svecs_left, svals, svecs_right = Primme.svds(A, 3, tol=1e-6, which='SM')
>>> svals # the three smallest singular values of A
array([ 1., 2., 3.])
>>> import Primme, scipy.sparse
>>> A = scipy.sparse.rand(10000, 100, random_state=10)
>>> prec = scipy.sparse.spdiags(np.reciprocal(A.multiply(A).sum(axis=0)),
... [0], 100, 100) # square diag. preconditioner
>>> svecs_left, svals, svecs_right = Primme.svds(A, 3, which=6.0, tol=1e-6, precAHA=prec)
>>> ["%.5f" % x for x in svals.flat] # the three closest singular values of A to 0.5
['5.99871', '5.99057', '6.01065']
"""
A = aslinearoperator(A)
m, n = A.shape
if k <= 0 or k > min(n, m):
raise ValueError("k=%d must be between 1 and min(A.shape)=%d" %
(k, min(n, m)))
if precAHA is not None:
precAHA = aslinearoperator(precAHA)
if precAHA.shape[0] != precAHA.shape[1] or precAHA.shape[0] != n:
raise ValueError('precAHA: expected square matrix with size %d' %
n)
if precAAH is not None:
precAAH = aslinearoperator(precAAH)
if precAAH.shape[0] != precAAH.shape[1] or precAAH.shape[0] != m:
raise ValueError('precAAH: expected square matrix with size %d' %
m)
if precAug is not None:
precAug = aslinearoperator(precAug)
if precAug.shape[0] != precAug.shape[1] or precAug.shape[0] != m + n:
raise ValueError('precAug: expected square matrix with size %d' %
(m + n))
class PSP(PrimmeSvdsParams):
def __init__(self):
PrimmeSvdsParams.__init__(self)
def matvec(self, X, transpose):
if transpose == 0:
return A.matmat(X)
else:
return A.H.matmat(X)
def prevec(self, X, mode):
if mode == primme_svds_op_AtA and precAHA is not None:
return precAHA.matmat(X)
elif mode == primme_svds_op_AAt and precAAH is not None:
return precAAH.matmat(X)
elif mode == primme_svds_op_augmented and precAug is not None:
return precAug.matmat(X)
return X
pp = PSP()
pp.m = A.shape[0]
pp.n = A.shape[1]
pp.numSvals = k
if which == 'LM':
pp.target = primme_svds_largest
elif which == 'SM':
pp.target = primme_svds_smallest
else:
try:
which = float(which)
except:
raise ValueError("which must be either 'LM', 'SM' or a number.")
pp.target = primme_svds_closest_abs
pp.targetShifts = np.array([which], dtype='d')
pp.eps = tol
if ncv:
pp.maxBasisSize = ncv
if maxiter:
# NOTE: every eigensolver iteration spend two matvecs*blockSize
pp.maxMatvecs = maxiter * (maxBlockSize if maxBlockSize else 1) / 2
if maxBlockSize:
pp.maxBlockSize = maxBlockSize
if precAHA is not None or precAAH is not None or precAug is not None:
pp.precondition = 1
def check_pair(u, v, var_names):
if ((u is not None and u.shape[0] != m)
or (v is not None and v.shape[0] != n)):
aux = v
v = u
u = aux
if ((u is not None and u.shape[0] != m)
or (v is not None and v.shape[0] != n)):
aux = v
v = u
u = aux
raise ValueError("%s don't have the expected number of rows." %
var_names)
if u is not None and v is not None and u.shape[1] != v.shape[1]:
raise ValueError("%s don't have the same number of columns." %
var_names)
if u is not None and v is None:
v, _ = np.linalg.qr(A.H.matmult(u))
if v is not None and u is None:
u, _ = np.linalg.qr(A.matmult(v))
return u, v
locku0, lockv0 = check_pair(locku0, lockv0, "lockv0 or locku0")
if locku0 is not None:
pp.numOrthoConst = min(locku0.shape[1], min(m, n))
if A.dtype.kind in frozenset(["b", "i", "u", "f"]):
dtype = np.dtype("d")
else:
dtype = np.dtype("complex128")
svals = np.zeros(pp.numSvals)
svecsl = np.zeros((pp.m, pp.numOrthoConst + pp.numSvals), dtype, order='F')
svecsr = np.zeros((pp.n, pp.numOrthoConst + pp.numSvals), dtype, order='F')
norms = np.zeros(pp.numSvals)
if locku0 is not None:
np.copyto(svecsl[:, 0:pp.numOrthoConst], locku0[:, 0:pp.numOrthoConst])
np.copyto(svecsr[:, 0:pp.numOrthoConst], lockv0[:, 0:pp.numOrthoConst])
u0, v0 = check_pair(u0, v0, "v0 or u0")
if v0 is not None:
pp.initSize = min(v0.shape[1], pp.numSvals)
np.copyto(svecsl[:, pp.numOrthoConst:pp.numOrthoConst + pp.initSize],
u0[:, 0:pp.initSize])
np.copyto(svecsr[:, pp.numOrthoConst:pp.numOrthoConst + pp.initSize],
v0[:, 0:pp.initSize])
if dtype is np.dtype('d'):
err = dprimme_svds(svals, svecsl, svecsr, norms, pp)
else:
err = zprimme_svds(svals, svecsl, svecsr, norms, pp)
if err != 0:
raise PrimmeSvdsError(err)
if return_stats:
stats = dict((f, getattr(pp.stats, f)) for f in [
"numOuterIterations", "numRestarts", "numMatvecs", "numPreconds",
"elapsedTime"
])
if not return_singular_vectors:
return svals if not return_stats else (svals, stats)
svecsl = svecsl[:, pp.numOrthoConst:]
svecsr = svecsr[:, pp.numOrthoConst:]
# Transpose conjugate svecsr
svecsr = svecsr.T.conj()
if not return_stats:
return svecsl, svals, svecsr
else:
return svecsl, svals, svecsr, stats
def lsqr(A, b, damp=0.0, atol=1e-8, btol=1e-8, conlim=1e8,
iter_lim=None, show=False, calc_var=False):
"""Find the least-squares solution to a large, sparse, linear system
of equations.
The function solves ``Ax = b`` or ``min ||b - Ax||^2`` or
``min ||Ax - b||^2 + d^2 ||x||^2``.
The matrix A may be square or rectangular (over-determined or
under-determined), and may have any rank.
::
1. Unsymmetric equations -- solve A*x = b
2. Linear least squares -- solve A*x = b
in the least-squares sense
3. Damped least squares -- solve ( A )*x = ( b )
( damp*I ) ( 0 )
in the least-squares sense
Parameters
----------
A : {sparse matrix, ndarray, LinearOperator}
Representation of an m-by-n matrix. It is required that
the linear operator can produce ``Ax`` and ``A^T x``.
b : (m,) ndarray
Right-hand side vector ``b``.
damp : float
Damping coefficient.
atol, btol : float, optional
Stopping tolerances. If both are 1.0e-9 (say), the final
residual norm should be accurate to about 9 digits. (The
final x will usually have fewer correct digits, depending on
cond(A) and the size of damp.)
conlim : float, optional
Another stopping tolerance. lsqr terminates if an estimate of
``cond(A)`` exceeds `conlim`. For compatible systems ``Ax =
b``, `conlim` could be as large as 1.0e+12 (say). For
least-squares problems, conlim should be less than 1.0e+8.
Maximum precision can be obtained by setting ``atol = btol =
conlim = zero``, but the number of iterations may then be
excessive.
iter_lim : int, optional
Explicit limitation on number of iterations (for safety).
show : bool, optional
Display an iteration log.
calc_var : bool, optional
Whether to estimate diagonals of ``(A'A + damp^2*I)^{-1}``.
Returns
-------
x : ndarray of float
The final solution.
istop : int
Gives the reason for termination.
1 means x is an approximate solution to Ax = b.
2 means x approximately solves the least-squares problem.
itn : int
Iteration number upon termination.
r1norm : float
``norm(r)``, where ``r = b - Ax``.
r2norm : float
``sqrt( norm(r)^2 + damp^2 * norm(x)^2 )``. Equal to `r1norm` if
``damp == 0``.
anorm : float
Estimate of Frobenius norm of ``Abar = [[A]; [damp*I]]``.
acond : float
Estimate of ``cond(Abar)``.
arnorm : float
Estimate of ``norm(A'*r - damp^2*x)``.
xnorm : float
``norm(x)``
var : ndarray of float
If ``calc_var`` is True, estimates all diagonals of
``(A'A)^{-1}`` (if ``damp == 0``) or more generally ``(A'A +
damp^2*I)^{-1}``. This is well defined if A has full column
rank or ``damp > 0``. (Not sure what var means if ``rank(A)
< n`` and ``damp = 0.``)
Notes
-----
LSQR uses an iterative method to approximate the solution. The
number of iterations required to reach a certain accuracy depends
strongly on the scaling of the problem. Poor scaling of the rows
or columns of A should therefore be avoided where possible.
For example, in problem 1 the solution is unaltered by
row-scaling. If a row of A is very small or large compared to
the other rows of A, the corresponding row of ( A b ) should be
scaled up or down.
In problems 1 and 2, the solution x is easily recovered
following column-scaling. Unless better information is known,
the nonzero columns of A should be scaled so that they all have
the same Euclidean norm (e.g., 1.0).
In problem 3, there is no freedom to re-scale if damp is
nonzero. However, the value of damp should be assigned only
after attention has been paid to the scaling of A.
The parameter damp is intended to help regularize
ill-conditioned systems, by preventing the true solution from
being very large. Another aid to regularization is provided by
the parameter acond, which may be used to terminate iterations
before the computed solution becomes very large.
If some initial estimate ``x0`` is known and if ``damp == 0``,
one could proceed as follows:
1. Compute a residual vector ``r0 = b - A*x0``.
2. Use LSQR to solve the system ``A*dx = r0``.
3. Add the correction dx to obtain a final solution ``x = x0 + dx``.
This requires that ``x0`` be available before and after the call
to LSQR. To judge the benefits, suppose LSQR takes k1 iterations
to solve A*x = b and k2 iterations to solve A*dx = r0.
If x0 is "good", norm(r0) will be smaller than norm(b).
If the same stopping tolerances atol and btol are used for each
system, k1 and k2 will be similar, but the final solution x0 + dx
should be more accurate. The only way to reduce the total work
is to use a larger stopping tolerance for the second system.
If some value btol is suitable for A*x = b, the larger value
btol*norm(b)/norm(r0) should be suitable for A*dx = r0.
Preconditioning is another way to reduce the number of iterations.
If it is possible to solve a related system ``M*x = b``
efficiently, where M approximates A in some helpful way (e.g. M -
A has low rank or its elements are small relative to those of A),
LSQR may converge more rapidly on the system ``A*M(inverse)*z =
b``, after which x can be recovered by solving M*x = z.
If A is symmetric, LSQR should not be used!
Alternatives are the symmetric conjugate-gradient method (cg)
and/or SYMMLQ. SYMMLQ is an implementation of symmetric cg that
applies to any symmetric A and will converge more rapidly than
LSQR. If A is positive definite, there are other implementations
of symmetric cg that require slightly less work per iteration than
SYMMLQ (but will take the same number of iterations).
References
----------
.. [1] C. C. Paige and M. A. Saunders (1982a).
"LSQR: An algorithm for sparse linear equations and
sparse least squares", ACM TOMS 8(1), 43-71.
.. [2] C. C. Paige and M. A. Saunders (1982b).
"Algorithm 583. LSQR: Sparse linear equations and least
squares problems", ACM TOMS 8(2), 195-209.
.. [3] M. A. Saunders (1995). "Solution of sparse rectangular
systems using LSQR and CRAIG", BIT 35, 588-604.
"""
A = aslinearoperator(A)
if len(b.shape) > 1:
b = b.squeeze()
m, n = A.shape
if iter_lim is None:
iter_lim = 2 * n
var = np.zeros(n)
msg = ('The exact solution is x = 0 ',
'Ax - b is small enough, given atol, btol ',
'The least-squares solution is good enough, given atol ',
'The estimate of cond(Abar) has exceeded conlim ',
'Ax - b is small enough for this machine ',
'The least-squares solution is good enough for this machine',
'Cond(Abar) seems to be too large for this machine ',
'The iteration limit has been reached ')
if show:
print(' ')
print('LSQR Least-squares solution of Ax = b')
str1 = 'The matrix A has %8g rows and %8g cols' % (m, n)
str2 = 'damp = %20.14e calc_var = %8g' % (damp, calc_var)
str3 = 'atol = %8.2e conlim = %8.2e' % (atol, conlim)
str4 = 'btol = %8.2e iter_lim = %8g' % (btol, iter_lim)
print(str1)
print(str2)
print(str3)
print(str4)
itn = 0
istop = 0
ctol = 0
if conlim > 0:
ctol = 1/conlim
anorm = 0
acond = 0
dampsq = damp**2
ddnorm = 0
res2 = 0
xnorm = 0
xxnorm = 0
z = 0
cs2 = -1
sn2 = 0
"""
Set up the first vectors u and v for the bidiagonalization.
These satisfy beta*u = b, alfa*v = A'u.
"""
v = np.zeros(n)
u = b
x = np.zeros(n)
alfa = 0
beta = np.linalg.norm(u)
w = np.zeros(n)
if beta > 0:
u = (1/beta) * u
v = A.rmatvec(u)
alfa = np.linalg.norm(v)
if alfa > 0:
v = (1/alfa) * v
w = v.copy()
rhobar = alfa
phibar = beta
bnorm = beta
rnorm = beta
r1norm = rnorm
r2norm = rnorm
# Reverse the order here from the original matlab code because
# there was an error on return when arnorm==0
arnorm = alfa * beta
if arnorm == 0:
print(msg[0])
return x, istop, itn, r1norm, r2norm, anorm, acond, arnorm, xnorm, var
head1 = ' Itn x[0] r1norm r2norm '
head2 = ' Compatible LS Norm A Cond A'
if show:
print(' ')
print(head1, head2)
test1 = 1
test2 = alfa / beta
str1 = '%6g %12.5e' % (itn, x[0])
str2 = ' %10.3e %10.3e' % (r1norm, r2norm)
str3 = ' %8.1e %8.1e' % (test1, test2)
print(str1, str2, str3)
# Main iteration loop.
while itn < iter_lim:
itn = itn + 1
"""
% Perform the next step of the bidiagonalization to obtain the
% next beta, u, alfa, v. These satisfy the relations
% beta*u = a*v - alfa*u,
% alfa*v = A'*u - beta*v.
"""
u = A.matvec(v) - alfa * u
beta = np.linalg.norm(u)
if beta > 0:
u = (1/beta) * u
anorm = sqrt(anorm**2 + alfa**2 + beta**2 + damp**2)
v = A.rmatvec(u) - beta * v
alfa = np.linalg.norm(v)
if alfa > 0:
v = (1 / alfa) * v
# Use a plane rotation to eliminate the damping parameter.
# This alters the diagonal (rhobar) of the lower-bidiagonal matrix.
rhobar1 = sqrt(rhobar**2 + damp**2)
cs1 = rhobar / rhobar1
sn1 = damp / rhobar1
psi = sn1 * phibar
phibar = cs1 * phibar
# Use a plane rotation to eliminate the subdiagonal element (beta)
# of the lower-bidiagonal matrix, giving an upper-bidiagonal matrix.
cs, sn, rho = _sym_ortho(rhobar1, beta)
theta = sn * alfa
rhobar = -cs * alfa
phi = cs * phibar
phibar = sn * phibar
tau = sn * phi
# Update x and w.
t1 = phi / rho
t2 = -theta / rho
dk = (1 / rho) * w
x = x + t1 * w
w = v + t2 * w
ddnorm = ddnorm + np.linalg.norm(dk)**2
if calc_var:
var = var + dk**2
# Use a plane rotation on the right to eliminate the
# super-diagonal element (theta) of the upper-bidiagonal matrix.
# Then use the result to estimate norm(x).
delta = sn2 * rho
gambar = -cs2 * rho
rhs = phi - delta * z
zbar = rhs / gambar
xnorm = sqrt(xxnorm + zbar**2)
gamma = sqrt(gambar**2 + theta**2)
cs2 = gambar / gamma
sn2 = theta / gamma
z = rhs / gamma
xxnorm = xxnorm + z**2
# Test for convergence.
# First, estimate the condition of the matrix Abar,
# and the norms of rbar and Abar'rbar.
acond = anorm * sqrt(ddnorm)
res1 = phibar**2
res2 = res2 + psi**2
rnorm = sqrt(res1 + res2)
arnorm = alfa * abs(tau)
# Distinguish between
# r1norm = ||b - Ax|| and
# r2norm = rnorm in current code
# = sqrt(r1norm^2 + damp^2*||x||^2).
# Estimate r1norm from
# r1norm = sqrt(r2norm^2 - damp^2*||x||^2).
# Although there is cancellation, it might be accurate enough.
r1sq = rnorm**2 - dampsq * xxnorm
r1norm = sqrt(abs(r1sq))
if r1sq < 0:
r1norm = -r1norm
r2norm = rnorm
# Now use these norms to estimate certain other quantities,
# some of which will be small near a solution.
test1 = rnorm / bnorm
test2 = arnorm / (anorm * rnorm + eps)
test3 = 1 / (acond + eps)
t1 = test1 / (1 + anorm * xnorm / bnorm)
rtol = btol + atol * anorm * xnorm / bnorm
# The following tests guard against extremely small values of
# atol, btol or ctol. (The user may have set any or all of
# the parameters atol, btol, conlim to 0.)
# The effect is equivalent to the normal tests using
# atol = eps, btol = eps, conlim = 1/eps.
if itn >= iter_lim:
istop = 7
if 1 + test3 <= 1:
istop = 6
if 1 + test2 <= 1:
istop = 5
if 1 + t1 <= 1:
istop = 4
# Allow for tolerances set by the user.
if test3 <= ctol:
istop = 3
if test2 <= atol:
istop = 2
if test1 <= rtol:
istop = 1
# See if it is time to print something.
prnt = False
if n <= 40:
prnt = True
if itn <= 10:
prnt = True
if itn >= iter_lim-10:
prnt = True
# if itn%10 == 0: prnt = True
if test3 <= 2*ctol:
prnt = True
if test2 <= 10*atol:
prnt = True
if test1 <= 10*rtol:
prnt = True
if istop != 0:
prnt = True
if prnt:
if show:
str1 = '%6g %12.5e' % (itn, x[0])
str2 = ' %10.3e %10.3e' % (r1norm, r2norm)
str3 = ' %8.1e %8.1e' % (test1, test2)
str4 = ' %8.1e %8.1e' % (anorm, acond)
print(str1, str2, str3, str4)
if istop != 0:
break
# End of iteration loop.
# Print the stopping condition.
if show:
print(' ')
print('LSQR finished')
print(msg[istop])
print(' ')
str1 = 'istop =%8g r1norm =%8.1e' % (istop, r1norm)
str2 = 'anorm =%8.1e arnorm =%8.1e' % (anorm, arnorm)
str3 = 'itn =%8g r2norm =%8.1e' % (itn, r2norm)
str4 = 'acond =%8.1e xnorm =%8.1e' % (acond, xnorm)
print(str1 + ' ' + str2)
print(str3 + ' ' + str4)
print(' ')
return x, istop, itn, r1norm, r2norm, anorm, acond, arnorm, xnorm, var
def lsmr(A,
b,
damp=0.0,
atol=1e-6,
btol=1e-6,
conlim=1e8,
maxiter=None,
show=False,
x0=None):
"""Iterative solver for least-squares problems.
lsmr solves the system of linear equations ``Ax = b``. If the system
is inconsistent, it solves the least-squares problem ``min ||b - Ax||_2``.
``A`` is a rectangular matrix of dimension m-by-n, where all cases are
allowed: m = n, m > n, or m < n. ``b`` is a vector of length m.
The matrix A may be dense or sparse (usually sparse).
Parameters
----------
A : {matrix, sparse matrix, ndarray, LinearOperator}
Matrix A in the linear system.
Alternatively, ``A`` can be a linear operator which can
produce ``Ax`` and ``A^H x`` using, e.g.,
``scipy.sparse.linalg.LinearOperator``.
b : array_like, shape (m,)
Vector ``b`` in the linear system.
damp : float
Damping factor for regularized least-squares. `lsmr` solves
the regularized least-squares problem::
min ||(b) - ( A )x||
||(0) (damp*I) ||_2
where damp is a scalar. If damp is None or 0, the system
is solved without regularization.
atol, btol : float, optional
Stopping tolerances. `lsmr` continues iterations until a
certain backward error estimate is smaller than some quantity
depending on atol and btol. Let ``r = b - Ax`` be the
residual vector for the current approximate solution ``x``.
If ``Ax = b`` seems to be consistent, ``lsmr`` terminates
when ``norm(r) <= atol * norm(A) * norm(x) + btol * norm(b)``.
Otherwise, lsmr terminates when ``norm(A^H r) <=
atol * norm(A) * norm(r)``. If both tolerances are 1.0e-6 (say),
the final ``norm(r)`` should be accurate to about 6
digits. (The final ``x`` will usually have fewer correct digits,
depending on ``cond(A)`` and the size of LAMBDA.) If `atol`
or `btol` is None, a default value of 1.0e-6 will be used.
Ideally, they should be estimates of the relative error in the
entries of ``A`` and ``b`` respectively. For example, if the entries
of ``A`` have 7 correct digits, set ``atol = 1e-7``. This prevents
the algorithm from doing unnecessary work beyond the
uncertainty of the input area_data.
conlim : float, optional
`lsmr` terminates if an estimate of ``cond(A)`` exceeds
`conlim`. For compatible systems ``Ax = b``, conlim could be
as large as 1.0e+12 (say). For least-squares problems,
`conlim` should be less than 1.0e+8. If `conlim` is None, the
default value is 1e+8. Maximum precision can be obtained by
setting ``atol = btol = conlim = 0``, but the number of
iterations may then be excessive.
maxiter : int, optional
`lsmr` terminates if the number of iterations reaches
`maxiter`. The default is ``maxiter = min(m, n)``. For
ill-conditioned systems, a larger value of `maxiter` may be
needed.
show : bool, optional
Print iterations logs if ``show=True``.
x0 : array_like, shape (n,), optional
Initial guess of ``x``, if None zeros are used.
.. versionadded:: 1.0.0
Returns
-------
x : ndarray of float
Least-square solution returned.
istop : int
istop gives the reason for stopping::
istop = 0 means x=0 is a solution. If x0 was given, then x=x0 is a
solution.
= 1 means x is an approximate solution to A*x = B,
according to atol and btol.
= 2 means x approximately solves the least-squares problem
according to atol.
= 3 means COND(A) seems to be greater than CONLIM.
= 4 is the same as 1 with atol = btol = eps (machine
precision)
= 5 is the same as 2 with atol = eps.
= 6 is the same as 3 with CONLIM = 1/eps.
= 7 means ITN reached maxiter before the other stopping
conditions were satisfied.
itn : int
Number of iterations used.
normr : float
``norm(b-Ax)``
normar : float
``norm(A^H (b - Ax))``
norma : float
``norm(A)``
conda : float
Condition number of A.
normx : float
``norm(x)``
Notes
-----
.. versionadded:: 0.11.0
References
----------
.. [1] D. C.-L. Fong and M. A. Saunders,
"LSMR: An iterative algorithm for sparse least-squares problems",
SIAM J. Sci. Comput., vol. 33, pp. 2950-2971, 2011.
:arxiv:`1006.0758`
.. [2] LSMR Software, https://web.stanford.edu/group/SOL/software/lsmr/
Examples
--------
>>> from scipy.sparse import csc_matrix
>>> from scipy.sparse.linalg import lsmr
>>> A = csc_matrix([[1., 0.], [1., 1.], [0., 1.]], dtype=float)
The first example has the trivial solution `[0, 0]`
>>> b = np.array([0., 0., 0.], dtype=float)
>>> x, istop, itn, normr = lsmr(A, b)[:4]
>>> istop
0
>>> x
array([ 0., 0.])
The stopping code `istop=0` returned indicates that a vector of zeros was
found as a solution. The returned solution `x` indeed contains `[0., 0.]`.
The next example has a non-trivial solution:
>>> b = np.array([1., 0., -1.], dtype=float)
>>> x, istop, itn, normr = lsmr(A, b)[:4]
>>> istop
1
>>> x
array([ 1., -1.])
>>> itn
1
>>> normr
4.440892098500627e-16
As indicated by `istop=1`, `lsmr` found a solution obeying the tolerance
limits. The given solution `[1., -1.]` obviously solves the equation. The
remaining return values include information about the number of iterations
(`itn=1`) and the remaining difference of left and right side of the solved
equation.
The final example demonstrates the behavior in the case where there is no
solution for the equation:
>>> b = np.array([1., 0.01, -1.], dtype=float)
>>> x, istop, itn, normr = lsmr(A, b)[:4]
>>> istop
2
>>> x
array([ 1.00333333, -0.99666667])
>>> A.dot(x)-b
array([ 0.00333333, -0.00333333, 0.00333333])
>>> normr
0.005773502691896255
`istop` indicates that the system is inconsistent and thus `x` is rather an
approximate solution to the corresponding least-squares problem. `normr`
contains the minimal distance that was found.
"""
A = aslinearoperator(A)
b = atleast_1d(b)
if b.ndim > 1:
b = b.squeeze()
msg = ('The exact solution is x = 0, or x = x0, if x0 was given ',
'Ax - b is small enough, given atol, btol ',
'The least-squares solution is good enough, given atol ',
'The estimate of cond(Abar) has exceeded conlim ',
'Ax - b is small enough for this machine ',
'The least-squares solution is good enough for this machine',
'Cond(Abar) seems to be too large for this machine ',
'The iteration limit has been reached ')
hdg1 = ' itn x(1) norm r norm Ar'
hdg2 = ' compatible LS norm A cond A'
pfreq = 20 # print frequency (for repeating the heading)
pcount = 0 # print counter
m, n = A.shape
# stores the num of singular values
minDim = min([m, n])
if maxiter is None:
maxiter = minDim
if x0 is None:
dtype = result_type(A, b, float)
else:
dtype = result_type(A, b, x0, float)
if show:
print(' ')
print('LSMR Least-squares solution of Ax = b\n')
print(f'The matrix A has {m} rows and {n} columns')
print('damp = %20.14e\n' % (damp))
print('atol = %8.2e conlim = %8.2e\n' % (atol, conlim))
print('btol = %8.2e maxiter = %8g\n' % (btol, maxiter))
u = b
normb = norm(b)
if x0 is None:
x = zeros(n, dtype)
beta = normb.copy()
else:
x = atleast_1d(x0)
u = u - A.matvec(x)
beta = norm(u)
if beta > 0:
u = (1 / beta) * u
v = A.rmatvec(u)
alpha = norm(v)
else:
v = zeros(n, dtype)
alpha = 0
if alpha > 0:
v = (1 / alpha) * v
# Initialize variables for 1st iteration.
itn = 0
zetabar = alpha * beta
alphabar = alpha
rho = 1
rhobar = 1
cbar = 1
sbar = 0
h = v.copy()
hbar = zeros(n, dtype)
# Initialize variables for estimation of ||r||.
betadd = beta
betad = 0
rhodold = 1
tautildeold = 0
thetatilde = 0
zeta = 0
d = 0
# Initialize variables for estimation of ||A|| and cond(A)
normA2 = alpha * alpha
maxrbar = 0
minrbar = 1e+100
normA = sqrt(normA2)
condA = 1
normx = 0
# Items for use in stopping rules, normb set earlier
istop = 0
ctol = 0
if conlim > 0:
ctol = 1 / conlim
normr = beta
# Reverse the order here from the original matlab code because
# there was an error on return when arnorm==0
normar = alpha * beta
if normar == 0:
if show:
print(msg[0])
return x, istop, itn, normr, normar, normA, condA, normx
if show:
print(' ')
print(hdg1, hdg2)
test1 = 1
test2 = alpha / beta
str1 = '%6g %12.5e' % (itn, x[0])
str2 = ' %10.3e %10.3e' % (normr, normar)
str3 = ' %8.1e %8.1e' % (test1, test2)
print(''.join([str1, str2, str3]))
# Main iteration loop.
while itn < maxiter:
itn = itn + 1
# Perform the next step of the bidiagonalization to obtain the
# next beta, u, alpha, v. These satisfy the relations
# beta*u = a*v - alpha*u,
# alpha*v = A'*u - beta*v.
u *= -alpha
u += A.matvec(v)
beta = norm(u)
if beta > 0:
u *= (1 / beta)
v *= -beta
v += A.rmatvec(u)
alpha = norm(v)
if alpha > 0:
v *= (1 / alpha)
# At this point, beta = beta_{k+1}, alpha = alpha_{k+1}.
# Construct rotation Qhat_{k,2k+1}.
chat, shat, alphahat = _sym_ortho(alphabar, damp)
# Use a plane rotation (Q_i) to turn B_i to R_i
rhoold = rho
c, s, rho = _sym_ortho(alphahat, beta)
thetanew = s * alpha
alphabar = c * alpha
# Use a plane rotation (Qbar_i) to turn R_i^T to R_i^bar
rhobarold = rhobar
zetaold = zeta
thetabar = sbar * rho
rhotemp = cbar * rho
cbar, sbar, rhobar = _sym_ortho(cbar * rho, thetanew)
zeta = cbar * zetabar
zetabar = -sbar * zetabar
# Update h, h_hat, x.
hbar *= -(thetabar * rho / (rhoold * rhobarold))
hbar += h
x += (zeta / (rho * rhobar)) * hbar
h *= -(thetanew / rho)
h += v
# Estimate of ||r||.
# Apply rotation Qhat_{k,2k+1}.
betaacute = chat * betadd
betacheck = -shat * betadd
# Apply rotation Q_{k,k+1}.
betahat = c * betaacute
betadd = -s * betaacute
# Apply rotation Qtilde_{k-1}.
# betad = betad_{k-1} here.
thetatildeold = thetatilde
ctildeold, stildeold, rhotildeold = _sym_ortho(rhodold, thetabar)
thetatilde = stildeold * rhobar
rhodold = ctildeold * rhobar
betad = -stildeold * betad + ctildeold * betahat
# betad = betad_k here.
# rhodold = rhod_k here.
tautildeold = (zetaold - thetatildeold * tautildeold) / rhotildeold
taud = (zeta - thetatilde * tautildeold) / rhodold
d = d + betacheck * betacheck
normr = sqrt(d + (betad - taud)**2 + betadd * betadd)
# Estimate ||A||.
normA2 = normA2 + beta * beta
normA = sqrt(normA2)
normA2 = normA2 + alpha * alpha
# Estimate cond(A).
maxrbar = max(maxrbar, rhobarold)
if itn > 1:
minrbar = min(minrbar, rhobarold)
condA = max(maxrbar, rhotemp) / min(minrbar, rhotemp)
# Test for convergence.
# Compute norms for convergence testing.
normar = abs(zetabar)
normx = norm(x)
# Now use these norms to estimate certain other quantities,
# some of which will be small near a solution.
test1 = normr / normb
if (normA * normr) != 0:
test2 = normar / (normA * normr)
else:
test2 = infty
test3 = 1 / condA
t1 = test1 / (1 + normA * normx / normb)
rtol = btol + atol * normA * normx / normb
# The following tests guard against extremely small values of
# atol, btol or ctol. (The user may have set any or all of
# the parameters atol, btol, conlim to 0.)
# The effect is equivalent to the normAl tests using
# atol = eps, btol = eps, conlim = 1/eps.
if itn >= maxiter:
istop = 7
if 1 + test3 <= 1:
istop = 6
if 1 + test2 <= 1:
istop = 5
if 1 + t1 <= 1:
istop = 4
# Allow for tolerances set by the user.
if test3 <= ctol:
istop = 3
if test2 <= atol:
istop = 2
if test1 <= rtol:
istop = 1
# See if it is time to print something.
if show:
if (n <= 40) or (itn <= 10) or (itn >= maxiter - 10) or \
(itn % 10 == 0) or (test3 <= 1.1 * ctol) or \
(test2 <= 1.1 * atol) or (test1 <= 1.1 * rtol) or \
(istop != 0):
if pcount >= pfreq:
pcount = 0
print(' ')
print(hdg1, hdg2)
pcount = pcount + 1
str1 = '%6g %12.5e' % (itn, x[0])
str2 = ' %10.3e %10.3e' % (normr, normar)
str3 = ' %8.1e %8.1e' % (test1, test2)
str4 = ' %8.1e %8.1e' % (normA, condA)
print(''.join([str1, str2, str3, str4]))
if istop > 0:
break
# Print the stopping condition.
if show:
print(' ')
print('LSMR finished')
print(msg[istop])
print('istop =%8g normr =%8.1e' % (istop, normr))
print(' normA =%8.1e normAr =%8.1e' % (normA, normar))
print('itn =%8g condA =%8.1e' % (itn, condA))
print(' normx =%8.1e' % (normx))
print(str1, str2)
print(str3, str4)
return x, istop, itn, normr, normar, normA, condA, normx
def _aslinearoperator_with_dtype(m):
m = aslinearoperator(m)
if not hasattr(m, 'dtype'):
x = np.zeros(m.shape[1])
m.dtype = (m*x).dtype
return m
def lsmr(A, b, damp=0.0, atol=1e-6, btol=1e-6, conlim=1e8,
maxiter=None, show=False):
"""Iterative solver for least-squares problems.
lsmr solves the system of linear equations ``Ax = b``. If the system
is inconsistent, it solves the least-squares problem ``min ||b - Ax||_2``.
A is a rectangular matrix of dimension m-by-n, where all cases are
allowed: m = n, m > n, or m < n. B is a vector of length m.
The matrix A may be dense or sparse (usually sparse).
.. versionadded:: 0.11.0
Parameters
----------
A : {matrix, sparse matrix, ndarray, LinearOperator}
Matrix A in the linear system.
b : (m,) ndarray
Vector b in the linear system.
damp : float
Damping factor for regularized least-squares. `lsmr` solves
the regularized least-squares problem::
min ||(b) - ( A )x||
||(0) (damp*I) ||_2
where damp is a scalar. If damp is None or 0, the system
is solved without regularization.
atol, btol : float
Stopping tolerances. `lsmr` continues iterations until a
certain backward error estimate is smaller than some quantity
depending on atol and btol. Let ``r = b - Ax`` be the
residual vector for the current approximate solution ``x``.
If ``Ax = b`` seems to be consistent, ``lsmr`` terminates
when ``norm(r) <= atol * norm(A) * norm(x) + btol * norm(b)``.
Otherwise, lsmr terminates when ``norm(A^{T} r) <=
atol * norm(A) * norm(r)``. If both tolerances are 1.0e-6 (say),
the final ``norm(r)`` should be accurate to about 6
digits. (The final x will usually have fewer correct digits,
depending on ``cond(A)`` and the size of LAMBDA.) If `atol`
or `btol` is None, a default value of 1.0e-6 will be used.
Ideally, they should be estimates of the relative error in the
entries of A and B respectively. For example, if the entries
of `A` have 7 correct digits, set atol = 1e-7. This prevents
the algorithm from doing unnecessary work beyond the
uncertainty of the input data.
conlim : float
`lsmr` terminates if an estimate of ``cond(A)`` exceeds
`conlim`. For compatible systems ``Ax = b``, conlim could be
as large as 1.0e+12 (say). For least-squares problems,
`conlim` should be less than 1.0e+8. If `conlim` is None, the
default value is 1e+8. Maximum precision can be obtained by
setting ``atol = btol = conlim = 0``, but the number of
iterations may then be excessive.
maxiter : int
`lsmr` terminates if the number of iterations reaches
`maxiter`. The default is ``maxiter = min(m, n)``. For
ill-conditioned systems, a larger value of `maxiter` may be
needed.
show : bool
Print iterations logs if ``show=True``.
Returns
-------
x : ndarray of float
Least-square solution returned.
istop : int
istop gives the reason for stopping::
istop = 0 means x=0 is a solution.
= 1 means x is an approximate solution to A*x = B,
according to atol and btol.
= 2 means x approximately solves the least-squares problem
according to atol.
= 3 means COND(A) seems to be greater than CONLIM.
= 4 is the same as 1 with atol = btol = eps (machine
precision)
= 5 is the same as 2 with atol = eps.
= 6 is the same as 3 with CONLIM = 1/eps.
= 7 means ITN reached maxiter before the other stopping
conditions were satisfied.
itn : int
Number of iterations used.
normr : float
``norm(b-Ax)``
normar : float
``norm(A^T (b - Ax))``
norma : float
``norm(A)``
conda : float
Condition number of A.
normx : float
``norm(x)``
References
----------
.. [1] D. C.-L. Fong and M. A. Saunders,
"LSMR: An iterative algorithm for sparse least-squares problems",
SIAM J. Sci. Comput., vol. 33, pp. 2950-2971, 2011.
http://arxiv.org/abs/1006.0758
.. [2] LSMR Software, http://www.stanford.edu/~clfong/lsmr.html
"""
A = aslinearoperator(A)
b = b.squeeze()
msg = ('The exact solution is x = 0 ',
'Ax - b is small enough, given atol, btol ',
'The least-squares solution is good enough, given atol ',
'The estimate of cond(Abar) has exceeded conlim ',
'Ax - b is small enough for this machine ',
'The least-squares solution is good enough for this machine',
'Cond(Abar) seems to be too large for this machine ',
'The iteration limit has been reached ')
hdg1 = ' itn x(1) norm r norm A''r'
hdg2 = ' compatible LS norm A cond A'
pfreq = 20 # print frequency (for repeating the heading)
pcount = 0 # print counter
m, n = A.shape
# stores the num of singular values
minDim = min([m, n])
if maxiter is None:
maxiter = minDim
if show:
print(' ')
print('LSMR Least-squares solution of Ax = b\n')
print('The matrix A has %8g rows and %8g cols' % (m, n))
print('damp = %20.14e\n' % (damp))
print('atol = %8.2e conlim = %8.2e\n' % (atol, conlim))
print('btol = %8.2e maxiter = %8g\n' % (btol, maxiter))
u = b
beta = norm(u)
v = zeros(n)
alpha = 0
if beta > 0:
u = (1 / beta) * u
v = A.rmatvec(u)
alpha = norm(v)
if alpha > 0:
v = (1 / alpha) * v
# Initialize variables for 1st iteration.
itn = 0
zetabar = alpha * beta
alphabar = alpha
rho = 1
rhobar = 1
cbar = 1
sbar = 0
h = v.copy()
hbar = zeros(n)
x = zeros(n)
# Initialize variables for estimation of ||r||.
betadd = beta
betad = 0
rhodold = 1
tautildeold = 0
thetatilde = 0
zeta = 0
d = 0
# Initialize variables for estimation of ||A|| and cond(A)
normA2 = alpha * alpha
maxrbar = 0
minrbar = 1e+100
normA = sqrt(normA2)
condA = 1
normx = 0
# Items for use in stopping rules.
normb = beta
istop = 0
ctol = 0
if conlim > 0:
ctol = 1 / conlim
normr = beta
# Reverse the order here from the original matlab code because
# there was an error on return when arnorm==0
normar = alpha * beta
if normar == 0:
if show:
print(msg[0])
return x, istop, itn, normr, normar, normA, condA, normx
if show:
print(' ')
print(hdg1, hdg2)
test1 = 1
test2 = alpha / beta
str1 = '%6g %12.5e' % (itn, x[0])
str2 = ' %10.3e %10.3e' % (normr, normar)
str3 = ' %8.1e %8.1e' % (test1, test2)
print(''.join([str1, str2, str3]))
# Main iteration loop.
while itn < maxiter:
itn = itn + 1
# Perform the next step of the bidiagonalization to obtain the
# next beta, u, alpha, v. These satisfy the relations
# beta*u = a*v - alpha*u,
# alpha*v = A'*u - beta*v.
u = A.matvec(v) - alpha * u
beta = norm(u)
if beta > 0:
u = (1 / beta) * u
v = A.rmatvec(u) - beta * v
alpha = norm(v)
if alpha > 0:
v = (1 / alpha) * v
# At this point, beta = beta_{k+1}, alpha = alpha_{k+1}.
# Construct rotation Qhat_{k,2k+1}.
chat, shat, alphahat = _sym_ortho(alphabar, damp)
# Use a plane rotation (Q_i) to turn B_i to R_i
rhoold = rho
c, s, rho = _sym_ortho(alphahat, beta)
thetanew = s*alpha
alphabar = c*alpha
# Use a plane rotation (Qbar_i) to turn R_i^T to R_i^bar
rhobarold = rhobar
zetaold = zeta
thetabar = sbar * rho
rhotemp = cbar * rho
cbar, sbar, rhobar = _sym_ortho(cbar * rho, thetanew)
zeta = cbar * zetabar
zetabar = - sbar * zetabar
# Update h, h_hat, x.
hbar = h - (thetabar * rho / (rhoold * rhobarold)) * hbar
x = x + (zeta / (rho * rhobar)) * hbar
h = v - (thetanew / rho) * h
# Estimate of ||r||.
# Apply rotation Qhat_{k,2k+1}.
betaacute = chat * betadd
betacheck = -shat * betadd
# Apply rotation Q_{k,k+1}.
betahat = c * betaacute
betadd = -s * betaacute
# Apply rotation Qtilde_{k-1}.
# betad = betad_{k-1} here.
thetatildeold = thetatilde
ctildeold, stildeold, rhotildeold = _sym_ortho(rhodold, thetabar)
thetatilde = stildeold * rhobar
rhodold = ctildeold * rhobar
betad = - stildeold * betad + ctildeold * betahat
# betad = betad_k here.
# rhodold = rhod_k here.
tautildeold = (zetaold - thetatildeold * tautildeold) / rhotildeold
taud = (zeta - thetatilde * tautildeold) / rhodold
d = d + betacheck * betacheck
normr = sqrt(d + (betad - taud)**2 + betadd * betadd)
# Estimate ||A||.
normA2 = normA2 + beta * beta
normA = sqrt(normA2)
normA2 = normA2 + alpha * alpha
# Estimate cond(A).
maxrbar = max(maxrbar, rhobarold)
if itn > 1:
minrbar = min(minrbar, rhobarold)
condA = max(maxrbar, rhotemp) / min(minrbar, rhotemp)
# Test for convergence.
# Compute norms for convergence testing.
normar = abs(zetabar)
normx = norm(x)
# Now use these norms to estimate certain other quantities,
# some of which will be small near a solution.
test1 = normr / normb
if (normA * normr) != 0:
test2 = normar / (normA * normr)
else:
test2 = infty
test3 = 1 / condA
t1 = test1 / (1 + normA * normx / normb)
rtol = btol + atol * normA * normx / normb
# The following tests guard against extremely small values of
# atol, btol or ctol. (The user may have set any or all of
# the parameters atol, btol, conlim to 0.)
# The effect is equivalent to the normAl tests using
# atol = eps, btol = eps, conlim = 1/eps.
if itn >= maxiter:
istop = 7
if 1 + test3 <= 1:
istop = 6
if 1 + test2 <= 1:
istop = 5
if 1 + t1 <= 1:
istop = 4
# Allow for tolerances set by the user.
if test3 <= ctol:
istop = 3
if test2 <= atol:
istop = 2
if test1 <= rtol:
istop = 1
# See if it is time to print something.
if show:
if (n <= 40) or (itn <= 10) or (itn >= maxiter - 10) or \
(itn % 10 == 0) or (test3 <= 1.1 * ctol) or \
(test2 <= 1.1 * atol) or (test1 <= 1.1 * rtol) or \
(istop != 0):
if pcount >= pfreq:
pcount = 0
print(' ')
print(hdg1, hdg2)
pcount = pcount + 1
str1 = '%6g %12.5e' % (itn, x[0])
str2 = ' %10.3e %10.3e' % (normr, normar)
str3 = ' %8.1e %8.1e' % (test1, test2)
str4 = ' %8.1e %8.1e' % (normA, condA)
print(''.join([str1, str2, str3, str4]))
if istop > 0:
break
# Print the stopping condition.
if show:
print(' ')
print('LSMR finished')
print(msg[istop])
print('istop =%8g normr =%8.1e' % (istop, normr))
print(' normA =%8.1e normAr =%8.1e' % (normA, normar))
print('itn =%8g condA =%8.1e' % (itn, condA))
print(' normx =%8.1e' % (normx))
print(str1, str2)
print(str3, str4)
return x, istop, itn, normr, normar, normA, condA, normx
def setup_method(self):
self.cases = []
def make_cases(original, dtype):
cases = []
cases.append((matrix(original, dtype=dtype), original))
cases.append((np.array(original, dtype=dtype), original))
cases.append((sparse.csr_matrix(original, dtype=dtype), original))
# Test default implementations of _adjoint and _rmatvec, which
# refer to each other.
def mv(x, dtype):
y = original.dot(x)
if len(x.shape) == 2:
y = y.reshape(-1, 1)
return y
def rmv(x, dtype):
return original.T.conj().dot(x)
class BaseMatlike(interface.LinearOperator):
args = ()
def __init__(self, dtype):
self.dtype = np.dtype(dtype)
self.shape = original.shape
def _matvec(self, x):
return mv(x, self.dtype)
class HasRmatvec(BaseMatlike):
args = ()
def _rmatvec(self, x):
return rmv(x, self.dtype)
class HasAdjoint(BaseMatlike):
args = ()
def _adjoint(self):
shape = self.shape[1], self.shape[0]
matvec = partial(rmv, dtype=self.dtype)
rmatvec = partial(mv, dtype=self.dtype)
return interface.LinearOperator(matvec=matvec,
rmatvec=rmatvec,
dtype=self.dtype,
shape=shape)
class HasRmatmat(HasRmatvec):
def _matmat(self, x):
return original.dot(x)
def _rmatmat(self, x):
return original.T.conj().dot(x)
cases.append((HasRmatvec(dtype), original))
cases.append((HasAdjoint(dtype), original))
cases.append((HasRmatmat(dtype), original))
return cases
original = np.array([[1, 2, 3], [4, 5, 6]])
self.cases += make_cases(original, np.int32)
self.cases += make_cases(original, np.float32)
self.cases += make_cases(original, np.float64)
self.cases += [(interface.aslinearoperator(M).T, A.T)
for M, A in make_cases(original.T, np.float64)]
self.cases += [(interface.aslinearoperator(M).H, A.T.conj())
for M, A in make_cases(original.T, np.float64)]
original = np.array([[1, 2j, 3j], [4j, 5j, 6]])
self.cases += make_cases(original, np.complex_)
self.cases += [(interface.aslinearoperator(M).T, A.T)
for M, A in make_cases(original.T, np.complex_)]
self.cases += [(interface.aslinearoperator(M).H, A.T.conj())
for M, A in make_cases(original.T, np.complex_)]
def make_system(A, M, x0, b, xtype=None):
"""Make a linear system Ax=b
Parameters
----------
A : LinearOperator
sparse or dense matrix (or any valid input to aslinearoperator)
M : {LinearOperator, Nones}
preconditioner
sparse or dense matrix (or any valid input to aslinearoperator)
x0 : {array_like, None}
initial guess to iterative method
b : array_like
right hand side
xtype : {'f', 'd', 'F', 'D', None}
dtype of the x vector
Returns
-------
(A, M, x, b, postprocess)
A : LinearOperator
matrix of the linear system
M : LinearOperator
preconditioner
x : rank 1 ndarray
initial guess
b : rank 1 ndarray
right hand side
postprocess : function
converts the solution vector to the appropriate
type and dimensions (e.g. (N,1) matrix)
"""
A_ = A
A = aslinearoperator(A)
if A.shape[0] != A.shape[1]:
raise ValueError('expected square matrix, but got shape=%s' %
(A.shape, ))
N = A.shape[0]
b = asanyarray(b)
if not (b.shape == (N, 1) or b.shape == (N, )):
raise ValueError('A and b have incompatible dimensions')
if b.dtype.char not in 'fdFD':
b = b.astype('d') # upcast non-FP types to double
def postprocess(x):
if isinstance(b, matrix):
x = asmatrix(x)
return x.reshape(b.shape)
if xtype is None:
if hasattr(A, 'dtype'):
xtype = A.dtype.char
else:
xtype = A.matvec(b).dtype.char
xtype = coerce(xtype, b.dtype.char)
else:
warn(
'Use of xtype argument is deprecated. '
'Use LinearOperator( ... , dtype=xtype) instead.',
DeprecationWarning)
if xtype == 0:
xtype = b.dtype.char
else:
if xtype not in 'fdFD':
raise ValueError("xtype must be 'f', 'd', 'F', or 'D'")
b = asarray(b, dtype=xtype) # make b the same type as x
b = b.ravel()
if x0 is None:
x = zeros(N, dtype=xtype)
else:
x = array(x0, dtype=xtype)
if not (x.shape == (N, 1) or x.shape == (N, )):
raise ValueError('A and x have incompatible dimensions')
x = x.ravel()
# process preconditioner
if M is None:
if hasattr(A_, 'psolve'):
psolve = A_.psolve
else:
psolve = id
if hasattr(A_, 'rpsolve'):
rpsolve = A_.rpsolve
else:
rpsolve = id
if psolve is id and rpsolve is id:
M = IdentityOperator(shape=A.shape, dtype=A.dtype)
else:
M = LinearOperator(A.shape,
matvec=psolve,
rmatvec=rpsolve,
dtype=A.dtype)
else:
M = aslinearoperator(M)
if A.shape != M.shape:
raise ValueError('matrix and preconditioner have different shapes')
return A, M, x, b, postprocess
def cgnr(A, b, x0=None, tol=1e-5, maxiter=None, xtype=None, M=None,
callback=None, residuals=None):
'''Conjugate Gradient, Normal Residual algorithm
Applies CG to the normal equations, A.H A x = b. Left preconditioning
is supported. Note that unless A is well-conditioned, the use of
CGNR is inadvisable
Parameters
----------
A : {array, matrix, sparse matrix, LinearOperator}
n x n, linear system to solve
b : {array, matrix}
right hand side, shape is (n,) or (n,1)
x0 : {array, matrix}
initial guess, default is a vector of zeros
tol : float
relative convergence tolerance, i.e. tol is scaled by ||r_0||_2
maxiter : int
maximum number of allowed iterations
xtype : type
dtype for the solution, default is automatic type detection
M : {array, matrix, sparse matrix, LinearOperator}
n x n, inverted preconditioner, i.e. solve M A.H A x = b.
callback : function
User-supplied function is called after each iteration as
callback(xk), where xk is the current solution vector
residuals : list
residuals has the residual norm history,
including the initial residual, appended to it
Returns
-------
(xNew, info)
xNew : an updated guess to the solution of Ax = b
info : halting status of cgnr
== =======================================
0 successful exit
>0 convergence to tolerance not achieved,
return iteration count instead.
<0 numerical breakdown, or illegal input
== =======================================
Notes
-----
The LinearOperator class is in scipy.sparse.linalg.interface.
Use this class if you prefer to define A or M as a mat-vec routine
as opposed to explicitly constructing the matrix. A.psolve(..) is
still supported as a legacy.
Examples
--------
>>> from pyamg.krylov.cgnr import cgnr
>>> from pyamg.util.linalg import norm
>>> import numpy as np
>>> from pyamg.gallery import poisson
>>> A = poisson((10,10))
>>> b = np.ones((A.shape[0],))
>>> (x,flag) = cgnr(A,b, maxiter=2, tol=1e-8)
>>> print norm(b - A*x)
9.3910201849
References
----------
.. [1] Yousef Saad, "Iterative Methods for Sparse Linear Systems,
Second Edition", SIAM, pp. 276-7, 2003
http://www-users.cs.umn.edu/~saad/books.html
'''
# Store the conjugate transpose explicitly as it will be used much later on
if isspmatrix(A):
AH = A.H
else:
# TODO avoid doing this since A may be a different sparse type
AH = aslinearoperator(asmatrix(A).H)
# Convert inputs to linear system, with error checking
A, M, x, b, postprocess = make_system(A, M, x0, b)
dimen = A.shape[0]
# Ensure that warnings are always reissued from this function
import warnings
warnings.filterwarnings('always', module='pyamg\.krylov\._cgnr')
# Choose type
if not hasattr(A, 'dtype'):
Atype = upcast(x.dtype, b.dtype)
else:
Atype = A.dtype
if not hasattr(M, 'dtype'):
Mtype = upcast(x.dtype, b.dtype)
else:
Mtype = M.dtype
xtype = upcast(Atype, x.dtype, b.dtype, Mtype)
# Should norm(r) be kept
if residuals == []:
keep_r = True
else:
keep_r = False
# How often should r be recomputed
recompute_r = 8
# Check iteration numbers. CGNR suffers from loss of orthogonality quite
# easily, so we arbitrarily let the method go up to 130% over the
# theoretically necessary limit of maxiter=dimen
if maxiter is None:
maxiter = int(ceil(1.3*dimen)) + 2
elif maxiter < 1:
raise ValueError('Number of iterations must be positive')
elif maxiter > (1.3*dimen):
warn('maximum allowed inner iterations (maxiter) are the 130% times \
the number of dofs')
maxiter = int(ceil(1.3*dimen)) + 2
# Prep for method
r = b - A*x
rhat = AH*r
normr = norm(r)
if keep_r:
residuals.append(normr)
# Check initial guess ( scaling by b, if b != 0,
# must account for case when norm(b) is very small)
normb = norm(b)
if normb == 0.0:
normb = 1.0
if normr < tol*normb:
if callback is not None:
callback(x)
return (postprocess(x), 0)
# Scale tol by ||r_0||_2
if normr != 0.0:
tol = tol*normr
# Begin CGNR
# Apply preconditioner and calculate initial search direction
z = M*rhat
p = z.copy()
old_zr = inner(z.conjugate(), rhat)
for iter in range(maxiter):
# w_j = A p_j
w = A*p
# alpha = (z_j, rhat_j) / (w_j, w_j)
alpha = old_zr / inner(w.conjugate(), w)
# x_{j+1} = x_j + alpha*p_j
x += alpha*p
# r_{j+1} = r_j - alpha*w_j
if mod(iter, recompute_r) and iter > 0:
r -= alpha*w
else:
r = b - A*x
# rhat_{j+1} = A.H*r_{j+1}
rhat = AH*r
# z_{j+1} = M*r_{j+1}
z = M*rhat
# beta = (z_{j+1}, rhat_{j+1}) / (z_j, rhat_j)
new_zr = inner(z.conjugate(), rhat)
beta = new_zr / old_zr
old_zr = new_zr
# p_{j+1} = A.H*z_{j+1} + beta*p_j
p *= beta
p += z
# Allow user access to residual
if callback is not None:
callback(x)
# test for convergence
normr = norm(r)
if keep_r:
residuals.append(normr)
if normr < tol:
return (postprocess(x), 0)
# end loop
return (postprocess(x), iter+1)
def assertCompatibleSystem(self, A, xtrue):
Afun = aslinearoperator(A)
b = Afun.matvec(xtrue)
x = lsmr(A,b)[0]
assert_almost_equal(norm(x - xtrue), 0, 6)
def eigen_symmetric(A, k=6, M=None, sigma=None, which='LM', v0=None,
ncv=None, maxiter=None, tol=0,
return_eigenvectors=True):
"""Find k eigenvalues and eigenvectors of the real symmetric
square matrix A.
Solves A * x[i] = w[i] * x[i], the standard eigenvalue problem for
w[i] eigenvalues with corresponding eigenvectors x[i].
Parameters
----------
A : matrix or array with real entries or object with matvec(x) method
An N x N real symmetric matrix or array or an object with matvec(x)
method to perform the matrix vector product A * x. The sparse
matrix formats in scipy.sparse are appropriate for A.
k : integer
The number of eigenvalues and eigenvectors desired
Returns
-------
w : array
Array of k eigenvalues
v : array
An array of k eigenvectors
The v[i] is the eigenvector corresponding to the eigenvector w[i]
Other Parameters
----------------
M : matrix or array
(Not implemented)
A symmetric positive-definite matrix for the generalized
eigenvalue problem A * x = w * M * x
sigma : real
(Not implemented)
Find eigenvalues near sigma. Shift spectrum by sigma.
v0 : array
Starting vector for iteration.
ncv : integer
The number of Lanczos vectors generated
ncv must be greater than k; it is recommended that ncv > 2*k
which : string
Which k eigenvectors and eigenvalues to find:
- 'LA' : Largest (algebraic) eigenvalues
- 'SA' : Smallest (algebraic) eigenvalues
- 'LM' : Largest (in magnitude) eigenvalues
- 'SM' : Smallest (in magnitude) eigenvalues
- 'BE' : Half (k/2) from each end of the spectrum
When k is odd, return one more (k/2+1) from the high end
maxiter : integer
Maximum number of Arnoldi update iterations allowed
tol : float
Relative accuracy for eigenvalues (stopping criterion)
return_eigenvectors : boolean
Return eigenvectors (True) in addition to eigenvalues
See Also
--------
eigen : eigenvalues and eigenvectors for a general (nonsymmetric) matrix A
Notes
-----
Examples
--------
"""
A = aslinearoperator(A)
if A.shape[0] != A.shape[1]:
raise ValueError('expected square matrix (shape=%s)' % shape)
n = A.shape[0]
# guess type
typ = A.dtype.char
if typ not in 'fd':
raise ValueError("matrix must be real valued (type must be 'f' or 'd')")
if M is not None:
raise NotImplementedError("generalized eigenproblem not supported yet")
if sigma is not None:
raise NotImplementedError("shifted eigenproblem not supported yet")
if ncv is None:
ncv=2*k+1
ncv=min(ncv,n)
if maxiter==None:
maxiter=n*10
# assign starting vector
if v0 is not None:
resid=v0
info=1
else:
resid = np.zeros(n,typ)
info=0
# some sanity checks
if k <= 0:
raise ValueError("k must be positive, k=%d"%k)
if k == n:
raise ValueError("k must be less than rank(A), k=%d"%k)
if maxiter <= 0:
raise ValueError("maxiter must be positive, maxiter=%d"%maxiter)
whiches=['LM','SM','LA','SA','BE']
if which not in whiches:
raise ValueError("which must be one of %s"%' '.join(whiches))
if ncv > n or ncv < k:
raise ValueError("ncv must be k<=ncv<=n, ncv=%s"%ncv)
# assign solver and postprocessor
ltr = _type_conv[typ]
eigsolver = _arpack.__dict__[ltr+'saupd']
eigextract = _arpack.__dict__[ltr+'seupd']
# set output arrays, parameters, and workspace
v = np.zeros((n,ncv),typ)
workd = np.zeros(3*n,typ)
workl = np.zeros(ncv*(ncv+8),typ)
iparam = np.zeros(11,'int')
ipntr = np.zeros(11,'int')
ido = 0
# set solver mode and parameters
# only supported mode is 1: Ax=lx
ishfts = 1
mode1 = 1
bmat='I'
iparam[0] = ishfts
iparam[2] = maxiter
iparam[6] = mode1
while True:
ido,resid,v,iparam,ipntr,info =\
eigsolver(ido,bmat,which,k,tol,resid,v,
iparam,ipntr,workd,workl,info)
xslice = slice(ipntr[0]-1, ipntr[0]-1+n)
yslice = slice(ipntr[1]-1, ipntr[1]-1+n)
if ido == -1:
# initialization
workd[yslice]=A.matvec(workd[xslice])
elif ido == 1:
# compute y=Ax
workd[yslice]=A.matvec(workd[xslice])
else:
break
if info < -1 :
raise RuntimeError("Error info=%d in arpack" % info)
return None
if info == 1:
warnings.warn("Maximum number of iterations taken: %s" % iparam[2])
if iparam[4] < k:
warnings.warn("Only %d/%d eigenvectors converged" % (iparam[4], k))
# now extract eigenvalues and (optionally) eigenvectors
rvec = return_eigenvectors
ierr = 0
howmny = 'A' # return all eigenvectors
sselect = np.zeros(ncv,'int') # unused
sigma = 0.0 # no shifts, not implemented
d,z,info =\
eigextract(rvec,howmny,sselect,sigma,
bmat,which, k,tol,resid,v,iparam[0:7],ipntr,
workd[0:2*n],workl,ierr)
if ierr != 0:
raise RuntimeError("Error info=%d in arpack"%info)
return None
if return_eigenvectors:
return d,z
return d
def testComplexB(self):
A = 4 * eye(self.n) + ones((self.n, self.n))
xtrue = transpose(arange(self.n, 0, -1) * (1 + 1j))
b = aslinearoperator(A).matvec(xtrue)
x = lsmr(A, b)[0]
assert norm(x - xtrue) == pytest.approx(0, abs=1e-5)
def lad(A,
b,
rho,
alpha=1.3,
verbose=True,
ret_hist=False,
abstol=1e-1,
reltol=1e-2,
no_iter=30,
lsmr_atol=1e-6,
lsmr_btol=1e-6,
lsmr_conlim=1e8,
lsmr_iter=1000):
"""
A port of the demonstration by Boyd, et al. in Matlab which was modified to
to use LSMR for the x-step. Solves the following problem via ADMM:
``minimize ||Ax - b||_1``
More information can be found in the paper linked at:
http://www.stanford.edu/~boyd/papers/distr_opt_stat_learning_admm.html
Parameters
----------
A : {sparse matrix, ndarray, LinearOperator}
Representation of an m-by-n matrix. It is required that
the linear operator can produce ``Ax`` and ``A^T x``.
b : array_like, shape (m,)
Right-hand side vector ``b``.
rho : scalar
The augmented Lagrangian parameter, which will need to be set based
upon the application. Typical values have been around 0.05.
alpha : scalar
The over-relaxation parameter, which should be above 1.0 and are
typically below 1.8.
verbose : bool
Print out the progress per iteration
ret_hist : bool
Return the history of ADMM and LSQR per iteration.
abstol : scalar
Absolute tolerance for the ADMM algorithm. Not used currently.
reltol : scalar
Relative tolerance for the ADMM algorithm. Not used currently.
no_iter : int
Maximum number of iterations of ADMM to be run.
lsmr_atol : scalar
The relative tolerance for ``A`` in the LSMR algorithm.
lsmr_btol : scalar
The relative tolerance for ``b`` in the LSMR algorithm.
lsmr_conlim : scalar
The condition limit for ``A`` in the LSMR algorithm.
lsmr_iter : int
Maximum number of iterations of LSMR to be run each ADMM iteration.
Parameters
----------
x : array_like, shape (n,)
The L1 norm estimate of ``x`` in ``Ax = b``
"""
A = aslinearoperator(A)
b = np.asarray(b)
m = A.shape[0]
n = A.shape[1]
x = np.zeros(n)
z = np.zeros(m)
u = np.zeros(m)
Ax = np.zeros(m)
if ret_hist:
history = dict()
history['x'] = np.zeros((no_iter, n))
history['r_norm'] = np.zeros(no_iter)
history['s_norm'] = np.zeros(no_iter)
history['eps_prim'] = np.zeros(no_iter)
history['eps_dual'] = np.zeros(no_iter)
history['objective'] = np.zeros(no_iter)
history['lsmr'] = dict()
history['lsmr']['flag'] = np.zeros(no_iter)
history['lsmr']['iter'] = np.zeros(no_iter)
history['lsmr']['normr'] = np.zeros(no_iter)
history['lsmr']['normar'] = np.zeros(no_iter)
history['lsmr']['norma'] = np.zeros(no_iter)
history['lsmr']['conda'] = np.zeros(no_iter)
history['lsmr']['normx'] = np.zeros(no_iter)
if verbose:
pass
# print '%3s\t%10s\t%10s\t%10s\t%10s\t%10s' % (
# 'iter', 'r norm', 'eps pri', 's norm', 'eps dual', 'objective')
for iter_no in range(no_iter):
q = b + z - u
# Perform LSMR on the residual and add it to x as suggested by scipy
# https://docs.scipy.org/doc/scipy-0.18.1/reference/generated/scipy.sparse.linalg.lsqr.html
# to perform a warm-start on the algorithm.
# TODO: Warm start was added in v1.0, so change this to use that
# feature
r0 = q - Ax
(dx, lsmr_flag, lsmr_iter, lsmr_normr, lsmr_normar, lsmr_norma,
lsmr_conda, lsmr_normx) = \
lsmr(A, r0, 0, lsmr_atol, lsmr_btol, lsmr_conlim, lsmr_iter)
x += dx
if ret_hist:
history['x'][iter_no, :] = x.copy()
history['lsmr']['flag'][iter_no] = lsmr_flag
history['lsmr']['iter'][iter_no] = lsmr_iter
history['lsmr']['normr'][iter_no] = lsmr_normr
history['lsmr']['normar'][iter_no] = lsmr_normar
history['lsmr']['norma'][iter_no] = lsmr_norma
history['lsmr']['conda'][iter_no] = lsmr_conda
history['lsmr']['normx'][iter_no] = lsmr_normx
zold = z
Ax = A.matvec(x)
Ax_hat = alpha * Ax + (1 - alpha) * (zold + b)
z = _shrinkage(Ax_hat - b + u, 1 / rho)
u = u + (Ax_hat - z - b)
# Calculate the iteration progress
objval = np.linalg.norm(z, 1)
r_norm = np.linalg.norm(Ax - z - b)
s_norm = np.linalg.norm(rho * A.rmatvec(zold - z))
# And the stopping criterion
eps_prim = abstol * np.sqrt(n) + \
reltol * np.max((np.linalg.norm(Ax),
np.linalg.norm(-z),
np.linalg.norm(b)))
eps_dual = abstol * np.sqrt(n) + \
reltol * np.linalg.norm(rho * A.rmatvec(u))
if ret_hist:
history['r_norm'][iter_no] = r_norm
history['s_norm'][iter_no] = s_norm
history['eps_prim'][iter_no] = eps_prim
history['eps_dual'][iter_no] = eps_dual
history['objective'][iter_no] = objval
if (r_norm < eps_prim) & (s_norm < eps_dual):
break
if verbose:
pass
# print '%3s\t%10s\t%10s\t%10s\t%10s\t%10s' % (
# iter_no, r_norm, eps_prim, s_norm, eps_dual, objval)
if ret_hist:
return x, history
else:
return x
def MinresQLP(A,
b,
rtol,
maxit,
M=None,
shift=None,
maxxnorm=None,
Acondlim=None,
TranCond=None,
show=False,
rnormvec=False):
A = aslinearoperator(A)
if shift is None:
shift = 0
if maxxnorm is None:
maxxnorm = 1e7
if Acondlim is None:
Acondlim = 1e15
if TranCond is None:
TranCond = 1e7
if rnormvec:
resvec = []
Aresvec = []
n = len(b)
b = b.reshape(n, 1)
r2 = b
r3 = r2
beta1 = norm(r2)
if M is None:
noprecon = True
pass
else:
noprecon = False
r3 = Precond(M, r2)
beta1 = r3.T.dot(r2) #teta
if beta1 < 0:
print('Error: "M" is indefinite!')
else:
beta1 = np.sqrt(beta1)
## Initialize
flag0 = -2
flag = -2
iters = 0
QLPiter = 0
beta = 0
tau = 0
taul = 0
phi = beta1
betan = beta1
gmin = 0
cs = -1
sn = 0
cr1 = -1
sr1 = 0
cr2 = -1
sr2 = 0
dltan = 0
eplnn = 0
gama = 0
gamal = 0
gamal2 = 0
eta = 0
etal = 0
etal2 = 0
vepln = 0
veplnl = 0
veplnl2 = 0
ul3 = 0
ul2 = 0
ul = 0
u = 0
rnorm = betan
xnorm = 0
xl2norm = 0
Axnorm = 0
Anorm = 0
Acond = 1
relres = rnorm / (beta1 + 1e-50)
x = np.zeros((n, 1))
w = np.zeros((n, 1))
wl = np.zeros((n, 1))
if rnormvec:
resvec = np.append(resvec, beta1)
msg = [
' beta2 = 0. b and x are eigenvectors ', # -1
' beta1 = 0. The exact solution is x = 0 ', # 0
' A solution to Ax = b found, given rtol ', # 1
' Min-length solution for singular LS problem, given rtol', # 2
' A solution to Ax = b found, given eps ', # 3
' Min-length solution for singular LS problem, given eps ', # 4
' x has converged to an eigenvector ', # 5
' xnorm has exceeded maxxnorm ', # 6
' Acond has exceeded Acondlim ', # 7
' The iteration limit was reached ', # 8
' Least-squares problem but no converged solution yet '
] # 9
if show:
print(' ')
print('Enter Minres-QLP: ')
print('Min-length solution of symmetric(singular)', end=' ')
print('(A-sI)x = b or min ||(A-sI)x - b||')
#||Ax - b|| is ||(A-sI)x - b|| if shift != 0 here
hstr1 = ' n = %8g ||Ax - b|| = %8.2e ' % (n, beta1)
hstr2 = 'shift = %8.2e rtol = %8g' % (shift, rtol)
hstr3 = 'maxit = %8g maxxnorm = %8.2e ' % (maxit, maxxnorm)
hstr4 = 'Acondlim = %8.2e TranCond = %8g' % (Acondlim, TranCond)
print(hstr1, hstr2)
print(hstr3, hstr4)
#b = 0 --> x = 0 skip the main loop
if beta1 == 0:
flag = 0
while flag == flag0 and iters < maxit:
#lanczos
iters += 1
betal = beta
beta = betan
v = r3 / beta
r3 = Ax(A, v)
if shift == 0:
pass
else:
r3 = r3 - shift * v
if iters > 1:
r3 = r3 - r1 * beta / betal
alfa = np.real(r3.T.dot(v))
r3 = r3 - r2 * alfa / beta
r1 = r2
r2 = r3
if noprecon:
betan = norm(r3)
if iters == 1:
if betan == 0:
if alfa == 0:
flag = 0
break
else:
flag = -1
x = b / alfa
break
else:
r3 = Precond(M, r2)
betan = r2.T.dot(r3)
if betan > 0:
betan = np.sqrt(betan)
else:
print('Error: "M" is indefinite or singular!')
pnorm = np.sqrt(betal**2 + alfa**2 + betan**2)
#previous left rotation Q_{k-1}
dbar = dltan
dlta = cs * dbar + sn * alfa
epln = eplnn
gbar = sn * dbar - cs * alfa
eplnn = sn * betan
dltan = -cs * betan
dlta_QLP = dlta
#current left plane rotation Q_k
gamal3 = gamal2
gamal2 = gamal
gamal = gama
cs, sn, gama = SymGivens(gbar, betan)
gama_tmp = gama
taul2 = taul
taul = tau
tau = cs * phi
Axnorm = np.sqrt(Axnorm**2 + tau**2)
phi = sn * phi
#previous right plane rotation P_{k-2,k}
if iters > 2:
veplnl2 = veplnl
etal2 = etal
etal = eta
dlta_tmp = sr2 * vepln - cr2 * dlta
veplnl = cr2 * vepln + sr2 * dlta
dlta = dlta_tmp
eta = sr2 * gama
gama = -cr2 * gama
#current right plane rotation P{k-1,k}
if iters > 1:
cr1, sr1, gamal = SymGivens(gamal, dlta)
vepln = sr1 * gama
gama = -cr1 * gama
#update xnorm
xnorml = xnorm
ul4 = ul3
ul3 = ul2
if iters > 2:
ul2 = (taul2 - etal2 * ul4 - veplnl2 * ul3) / gamal2
if iters > 1:
ul = (taul - etal * ul3 - veplnl * ul2) / gamal
xnorm_tmp = np.sqrt(xl2norm**2 + ul2**2 + ul**2)
if abs(gama) > np.finfo(np.double).tiny and xnorm_tmp < maxxnorm:
u = (tau - eta * ul2 - vepln * ul) / gama
if np.sqrt(xnorm_tmp**2 + u**2) > maxxnorm:
u = 0
flag = 6
else:
u = 0
flag = 9
xl2norm = np.sqrt(xl2norm**2 + ul2**2)
xnorm = np.sqrt(xl2norm**2 + ul**2 + u**2)
#update w&x
#Minres
if (Acond < TranCond) and flag != flag0 and QLPiter == 0:
wl2 = wl
wl = w
w = (v - epln * wl2 - dlta_QLP * wl) / gama_tmp
if xnorm < maxxnorm:
x += tau * w
else:
flag = 6
#Minres-QLP
else:
QLPiter += 1
if QLPiter == 1:
xl2 = np.zeros((n, 1))
if (iters > 1): # construct w_{k-3}, w_{k-2}, w_{k-1}
if iters > 3:
wl2 = gamal3 * wl2 + veplnl2 * wl + etal * w
if iters > 2:
wl = gamal_QLP * wl + vepln_QLP * w
w = gama_QLP * w
xl2 = x - wl * ul_QLP - w * u_QLP
if iters == 1:
wl2 = wl
wl = v * sr1
w = -v * cr1
elif iters == 2:
wl2 = wl
wl = w * cr1 + v * sr1
w = w * sr1 - v * cr1
else:
wl2 = wl
wl = w
w = wl2 * sr2 - v * cr2
wl2 = wl2 * cr2 + v * sr2
v = wl * cr1 + w * sr1
w = wl * sr1 - w * cr1
wl = v
xl2 = xl2 + wl2 * ul2
x = xl2 + wl * ul + w * u
#next right plane rotation P{k-1,k+1}
gamal_tmp = gamal
cr2, sr2, gamal = SymGivens(gamal, eplnn)
#transfering from Minres to Minres-QLP
gamal_QLP = gamal_tmp
#print('gamal_QLP=', gamal_QLP)
vepln_QLP = vepln
gama_QLP = gama
ul_QLP = ul
u_QLP = u
## Estimate various norms
abs_gama = abs(gama)
Anorml = Anorm
Anorm = max([Anorm, pnorm, gamal, abs_gama])
if iters == 1:
gmin = gama
gminl = gmin
elif iters > 1:
gminl2 = gminl
gminl = gmin
gmin = min([gminl2, gamal, abs_gama])
Acondl = Acond
Acond = Anorm / gmin
rnorml = rnorm
relresl = relres
if flag != 9:
rnorm = phi
relres = rnorm / (Anorm * xnorm + beta1)
rootl = np.sqrt(gbar**2 + dltan**2)
Arnorml = rnorml * rootl
relAresl = rootl / Anorm
## See if any of the stopping criteria are satisfied.
epsx = Anorm * xnorm * np.finfo(float).eps
if (flag == flag0) or (flag == 9):
t1 = 1 + relres
t2 = 1 + relAresl
if iters >= maxit:
flag = 8 #exit before maxit
if Acond >= Acondlim:
flag = 7 #Huge Acond
if xnorm >= maxxnorm:
flag = 6 #xnorm exceeded
if epsx >= beta1:
flag = 5 #x = eigenvector
if t2 <= 1:
flag = 4 #Accurate Least Square Solution
if t1 <= 1:
flag = 3 #Accurate Ax = b Solution
if relAresl <= rtol:
flag = 2 #Trustful Least Square Solution
if relres <= rtol:
flag = 1 #Trustful Ax = b Solution
if flag == 2 or flag == 4 or flag == 6 or flag == 7:
#possibly singular
iters = iters - 1
Acond = Acondl
rnorm = rnorml
relres = relresl
else:
if rnormvec:
resvec = np.append(resvec, rnorm)
Aresvec = np.append(Aresvec, Arnorml)
if show:
if iters % 10 - 1 == 0:
lstr = (' iter rnorm Arnorm relres ' +
'relAres Anorm Acond xnorm')
print(' ')
print(lstr)
if QLPiter == 1:
print('QLP', end='')
else:
print(' ', end='')
lstr1 = '%8g %8.2e ' % (iters - 1, rnorml)
lstr2 = '%8.2e %8.2e ' % (Arnorml, relresl)
lstr3 = '%8.2e %8.2e ' % (relAresl, Anorml)
lstr4 = '%8.2e %8.2e ' % (Acondl, xnorml)
print(lstr1, lstr2, lstr3, lstr4)
#exited the main loop
if QLPiter == 1:
print('QLP', end='')
else:
print(' ', end='')
Miter = iters - QLPiter
#final quantities
r1 = b - Ax(A, x) + shift * x
rnorm = norm(r1)
Arnorm = norm(Ax(A, r1) - shift * r1)
xnorm = norm(x)
relres = rnorm / (Anorm * xnorm + beta1)
relAres = 0
if rnorm > np.finfo(np.double).tiny:
relAres = Arnorm / (Anorm * rnorm)
if show:
if rnorm > np.finfo(np.double).tiny:
lstr1 = '%8g %8.2e ' % (iters, rnorm)
lstr2 = '%8.2eD %8.2e ' % (Arnorm, relres)
lstr3 = '%8.2eD %8.2e ' % (relAres, Anorm)
lstr4 = '%8.2e %8.2e ' % (Acond, xnorm)
print(lstr1, lstr2, lstr3, lstr4)
else:
lstr1 = '%8g %8.2e ' % (iters, rnorm)
lstr2 = '%8.2eD %8.2e ' % (Arnorm, relres)
lstr3 = ' %8.2e ' % (Anorm)
lstr4 = '%8.2e %8.2e ' % (Acond, xnorm)
print(lstr1, lstr2, lstr3, lstr4)
print(' ')
print('Exit Minres-QLP: ')
str1 = 'Flag = %8g %8s' % (flag, msg[int(flag + 1)])
str2 = 'Iter = %8g ' % (iters)
str3 = 'Minres = %8g Minres-QLP = %8g' % (Miter, QLPiter)
str4 = 'relres = %8.2e relAres = %8.2e ' % (relres, relAres)
str5 = 'rnorm = %8.2e Arnorm = %8.2e' % (rnorm, Arnorm)
str6 = 'Anorm = %8.2e Acond = %8.2e ' % (Anorm, Acond)
str7 = 'xnorm = %8.2e Axnorm = %8.2e' % (xnorm, Axnorm)
print(str1)
print(str2, str3)
print(str4, str5)
print(str6, str7)
if rnormvec:
Aresvec = np.append(Aresvec, Arnorm)
return (x, flag, iters, Miter, QLPiter, relres, relAres, Anorm, Acond,
xnorm, Axnorm, resvec, Aresvec)
return (x, flag, iters, Miter, QLPiter, relres, relAres, Anorm, Acond,
xnorm, Axnorm)