def svd_gesvd(a, full_matrices=True, compute_uv=True, check_finite=True):
"""svd with LAPACK's '#gesvd' (with # = d/z for float/complex).
Similar as :func:`numpy.linalg.svd`, but use LAPACK 'gesvd' driver.
Works only with 2D arrays.
Outer part is based on the code of `numpy.linalg.svd`.
Parameters
----------
a, full_matrices, compute_uv :
See :func:`numpy.linalg.svd` for details.
check_finite :
check whether input arrays contain 'NaN' or 'inf'.
Returns
-------
U, S, Vh : ndarray
See :func:`numpy.linalg.svd` for details.
"""
a, wrap = _makearray(a) # uses order='C'
_assertNoEmpty2d(a)
_assertRank2(a)
if check_finite:
_assertFinite(a)
M, N = a.shape
# determine types
t, result_t = _commonType(a)
# t = type for calculation, (for my numpy version) actually always one of {double, cdouble}
# result_t = one of {single, double, csingle, cdouble}
is_complex = isComplexType(t)
real_t = _realType(t) # real version of t with same precision
# copy: the array is destroyed
a = _fastCopyAndTranspose(
t, a) # casts a to t, copy and transpose (=change to order='F')
lapack_routine = _get_gesvd(t)
# allocate output space & options
if compute_uv:
if full_matrices:
nu = M
lvt = N
option = b'A'
else:
nu = min(N, M)
lvt = min(N, M)
option = b'S'
u = np.zeros((M, nu), t, order='F')
vt = np.zeros((lvt, N), t, order='F')
else:
option = b'N'
nu = 1
u = np.empty((1, 1), t, order='F')
vt = np.empty((1, 1), t, order='F')
s = np.zeros((min(N, M), ), real_t, order='F')
INFO = c_int(0)
m = c_int(M)
n = c_int(N)
lu = c_int(u.shape[0])
lvt = c_int(vt.shape[0])
work = np.zeros((1, ), t)
lwork = c_int(-1) # first call with lwork=-1
args = [option, option, m, n, a, m, s, u, lu, vt, lvt, work, lwork, INFO]
if is_complex:
# differnt call signature: additional array 'rwork' of fixed size
rwork = np.zeros((5 * min(N, M), ), real_t)
args.insert(-1, rwork)
lapack_routine(
*args) # first call: just calculate the required `work` size
if INFO.value < 0:
raise Exception('%d-th argument had an illegal value' % INFO.value)
if is_complex:
lwork = int(work[0].real)
else:
lwork = int(work[0])
work = np.zeros((lwork, ), t, order='F')
args[11] = work
args[12] = c_int(lwork)
lapack_routine(*args) # second call: the actual calculation
if INFO.value < 0:
raise Exception('%d-th argument had an illegal value' % INFO.value)
if INFO.value > 0:
raise LinAlgError("SVD did not converge with 'gesvd'")
s = s.astype(_realType(result_t))
if compute_uv:
u = u.astype(result_t) # no repeated transpose: used fortran order
vt = vt.astype(result_t)
return wrap(u), s, wrap(vt)
else:
return s
def controllability_indices(A, B, tol=None):
"""Computes the controllability indices for a controllable pair (A, B)
Controllability indices are defined as the maximum number of independent
columns per input column of B in the following sense: consider the Kalman
controllability matrix (widely known as Krylov sequence) ::
C = [B, AB, ..., A^(n-1)B]
We start testing (starting from the left-most) every column of this matrix
whether it is a linear combination of the previous columns. Obviously,
since C is (n x nm), there are many ways to pick a linearly independent
subset. We select a version from [1]_. If a new column is dependent
we delete that column and keep doing this until we are left with a
full-rank square matrix (this is guaranteed if (A, B) is controllable).
Then at some point, we are necessarily left with columns that are obtained
from different input columns ::
̅C = [b₁,b₂,b₃...,Ab₁,Ab₃,...,A²b₁,A²b₃,...,A⁽ᵏ⁻¹⁾b₃,...]
For example, it seems like Ab₂ is deleted due to dependence on the previous
columns to its left. It can be shown that the higher powers will also be
dependent and can be removed too. By reordering these columns, we combine
the terms that involve each bⱼ ::
̅C = [b₁,Ab₁,A²b₁,b₂,b₃,Ab₃,A²b₃,...,A⁽ᵏ⁻¹⁾b₃,...]
The controllability index associated with each input column is defined as
the number of columns per bⱼ appearing in the resulting matrix. Here, 3
for first input 1 for second and so on.
If B is not full rank then the index will be returned as 0 as that column
bⱼ will be dropped too.
Parameters
----------
A : ndarray
2D (n, n) real-valued array
B : ndarray
2D (n, m) real-valued array
tol : float
Tolerance value for the Arnoldi iteration to decide linear dependence.
By default it is `sqrt(eps)*n²`
Returns
-------
ind : ndarray
1D array that holds the computed controllability indices. The sum of
the values add up to `n` if (A, B) is controllable.
Notes
-----
Though internally not used, this function can also be used as a
controllability/observability test by summing up the resulting indices and
comparing to `n`.
References
----------
.. [1] : W.M. Wonham, "Linear Multivariable Control: A Geometric Approach",
3rd edition, 1985, Springer, ISBN:9780387960715
"""
a, _ = _makearray(A)
b, _ = _makearray(B)
_assertRank2(a, b)
_assertNoEmpty2d(a, b)
_assertNdSquareness(a)
n, m = b.shape
if a.shape[0] != b.shape[0]:
raise ValueError("A and B should have the same number of rows")
# FIXME: Tolerance is a bit arbitrary for now!!
tol = np.sqrt(np.spacing(1.)) * n**2 if tol is None else tol
# Will be populated below
remaining_cols = np.arange(m)
indices = np.zeros(m, dtype=int)
# Get the orthonormal columns of b first
q, r, p = qr(b, mode='economic', pivoting=True)
rank_b = sum(np.abs(np.diag(r)) > max(m, n) * np.spacing(norm(b, 2)))
remaining_cols = remaining_cols[p][:rank_b].tolist()
q = q[:, :rank_b]
indices[remaining_cols] += 1
w = np.empty((n, 1), dtype=float)
# Start the iteration - at most n-1 spins
for k in range(1, n):
# prepare new A @ Q test vectors
q_bank = a @ q[:, -len(remaining_cols):]
for ind, col in enumerate(remaining_cols.copy()):
w[:] = q_bank[:, [ind]]
for reorthogonalization in range(2):
w -= ((q.T @ w).T * q).sum(axis=1, keepdims=True)
normw = norm(w)
if normw <= tol:
remaining_cols.remove(col)
continue
else:
q = np.append(q, w / normw, axis=1)
indices[col] += 1
if len(remaining_cols) == 0:
break
return indices
def controllability_indices(A, B, tol=None):
"""Computes the controllability indices for a controllable pair (A, B)
Controllability indices are defined as the maximum number of independent
columns per input column of B in the following sense: consider the Kalman
controllability matrix (widely known as Krylov sequence) ::
C = [B, AB, ..., A^(n-1)B]
We start testing (starting from the left-most) every column of this matrix
whether it is a linear combination of the previous columns. Obviously,
since C is (n x nm), there are many ways to pick a linearly independent
subset. We select a version from [1]_. If a new column is dependent
we delete that column and keep doing this until we are left with a
full-rank square matrix (this is guaranteed if (A, B) is controllable).
Then at some point, we are necessarily left with columns that are obtained
from different input columns ::
̅C = [b₁,b₂,b₃...,Ab₁,Ab₃,...,A²b₁,A²b₃,...,A⁽ᵏ⁻¹⁾b₃,...]
For example, it seems like Ab₂ is deleted due to dependence on the previous
columns to its left. It can be shown that the higher powers will also be
dependent and can be removed too. By reordering these columns, we combine
the terms that involve each bⱼ ::
̅C = [b₁,Ab₁,A²b₁,b₂,b₃,Ab₃,A²b₃,...,A⁽ᵏ⁻¹⁾b₃,...]
The controllability index associated with each input column is defined as
the number of columns per bⱼ appearing in the resulting matrix. Here, 3
for first input 1 for second and so on.
If B is not full rank then the index will be returned as 0 as that column
bⱼ will be dropped too.
Parameters
----------
A : ndarray
2D (n, n) real-valued array
B : ndarray
2D (n, m) real-valued array
tol : float
Tolerance value for the Arnoldi iteration to decide linear dependence.
By default it is `sqrt(eps)*n²`
Returns
-------
ind : ndarray
1D array that holds the computed controllability indices. The sum of
the values add up to `n` if (A, B) is controllable.
Notes
-----
Though internally not used, this function can also be used as a
controllability/observability test by summing up the resulting indices and
comparing to `n`.
References
----------
.. [1] : W.M. Wonham, "Linear Multivariable Control: A Geometric Approach",
3rd edition, 1985, Springer, ISBN:9780387960715
"""
a, _ = _makearray(A)
b, _ = _makearray(B)
_assertRank2(a, b)
_assertNoEmpty2d(a, b)
_assertNdSquareness(a)
n, m = b.shape
if a.shape[0] != b.shape[0]:
raise ValueError("A and B should have the same number of rows")
# FIXME: Tolerance is a bit arbitrary for now!!
tol = np.sqrt(np.spacing(1.))*n**2 if tol is None else tol
# Will be populated below
remaining_cols = np.arange(m)
indices = np.zeros(m, dtype=int)
# Get the orthonormal columns of b first
q, r, p = qr(b, mode='economic', pivoting=True)
rank_b = sum(np.abs(np.diag(r)) > max(m, n)*np.spacing(norm(b, 2)))
remaining_cols = remaining_cols[p][:rank_b].tolist()
q = q[:, :rank_b]
indices[remaining_cols] += 1
w = np.empty((n, 1), dtype=float)
# Start the iteration - at most n-1 spins
for k in range(1, n):
# prepare new A @ Q test vectors
q_bank = a @ q[:, -len(remaining_cols):]
for ind, col in enumerate(remaining_cols.copy()):
w[:] = q_bank[:, [ind]]
for reorthogonalization in range(2):
w -= ((q.T @ w).T * q).sum(axis=1, keepdims=True)
normw = norm(w)
if normw <= tol:
remaining_cols.remove(col)
continue
else:
q = np.append(q, w/normw, axis=1)
indices[col] += 1
if len(remaining_cols) == 0:
break
return indices
def lstsq(a, b, rcond=-1):
"""
Return the least-squares solution to an equation.
Solves the equation `a x = b` by computing a vector `x` that minimizes
the norm `|| b - a x ||`.
Parameters
----------
a : array_like, shape (M, N)
Input equation coefficients.
b : array_like, shape (M,) or (M, K)
Equation target values. If `b` is two-dimensional, the least
squares solution is calculated for each of the `K` target sets.
rcond : float, optional
Cutoff for ``small`` singular values of `a`.
Singular values smaller than `rcond` times the largest singular
value are considered zero.
Returns
-------
x : ndarray, shape(N,) or (N, K)
Least squares solution. The shape of `x` depends on the shape of
`b`.
residues : ndarray, shape(), (1,), or (K,)
Sums of residues; squared Euclidian norm for each column in
`b - a x`.
If the rank of `a` is < N or > M, this is an empty array.
If `b` is 1-dimensional, this is a (1,) shape array.
Otherwise the shape is (K,).
rank : integer
Rank of matrix `a`.
s : ndarray, shape(min(M,N),)
Singular values of `a`.
Raises
------
LinAlgError
If computation does not converge.
Notes
-----
If `b` is a matrix, then all array results returned as
matrices.
Examples
--------
Fit a line, ``y = mx + c``, through some noisy data-points:
>>> x = np.array([0, 1, 2, 3])
>>> y = np.array([-1, 0.2, 0.9, 2.1])
By examining the coefficients, we see that the line should have a
gradient of roughly 1 and cuts the y-axis at more-or-less -1.
We can rewrite the line equation as ``y = Ap``, where ``A = [[x 1]]``
and ``p = [[m], [c]]``. Now use `lstsq` to solve for `p`:
>>> A = np.vstack([x, np.ones(len(x))]).T
>>> A
array([[ 0., 1.],
[ 1., 1.],
[ 2., 1.],
[ 3., 1.]])
>>> m, c = np.linalg.lstsq(A, y)[0]
>>> print m, c
1.0 -0.95
Plot the data along with the fitted line:
>>> import matplotlib.pyplot as plt
>>> plt.plot(x, y, 'o', label='Original data', markersize=10)
>>> plt.plot(x, m*x + c, 'r', label='Fitted line')
>>> plt.legend()
>>> plt.show()
"""
import math
a, _ = _makearray(a)
b, wrap = _makearray(b)
is_1d = len(b.shape) == 1
if is_1d:
b = b[:, newaxis]
_assertRank2(a, b)
m = a.shape[0]
n = a.shape[1]
n_rhs = b.shape[1]
ldb = max(n, m)
if m != b.shape[0]:
raise LinAlgError, 'Incompatible dimensions'
t, result_t = _commonType(a, b)
real_t = _linalgRealType(t)
bstar = zeros((ldb, n_rhs), t)
bstar[:b.shape[0],:n_rhs] = b.copy()
a, bstar = _fastCopyAndTranspose(t, a, bstar)
s = zeros((min(m, n),), real_t)
nlvl = max( 0, int( math.log( float(min(m, n))/2. ) ) + 1 )
iwork = zeros((3*min(m, n)*nlvl+11*min(m, n),), fortran_int)
if isComplexType(t):
lapack_routine = lapack_lite.zgelsd
lwork = 1
rwork = zeros((lwork,), real_t)
work = zeros((lwork,), t)
results = lapack_routine(m, n, n_rhs, a, m, bstar, ldb, s, rcond,
0, work, -1, rwork, iwork, 0)
lwork = int(abs(work[0]))
rwork = zeros((lwork,), real_t)
a_real = zeros((m, n), real_t)
bstar_real = zeros((ldb, n_rhs,), real_t)
results = lapack_lite.dgelsd(m, n, n_rhs, a_real, m,
bstar_real, ldb, s, rcond,
0, rwork, -1, iwork, 0)
lrwork = int(rwork[0])
work = zeros((lwork,), t)
rwork = zeros((lrwork,), real_t)
results = lapack_routine(m, n, n_rhs, a, m, bstar, ldb, s, rcond,
0, work, lwork, rwork, iwork, 0)
else:
lapack_routine = lapack_lite.dgelsd
lwork = 1
work = zeros((lwork,), t)
results = lapack_routine(m, n, n_rhs, a, m, bstar, ldb, s, rcond,
0, work, -1, iwork, 0)
lwork = int(work[0])
work = zeros((lwork,), t)
results = lapack_routine(m, n, n_rhs, a, m, bstar, ldb, s, rcond,
0, work, lwork, iwork, 0)
if results['info'] > 0:
raise LinAlgError, 'SVD did not converge in Linear Least Squares'
resids = array([], t)
if is_1d:
x = array(ravel(bstar)[:n], dtype=result_t, copy=True)
if results['rank'] == n and m > n:
resids = array([np.linalg.norm(ravel(bstar)[n:], 2)**2])
else:
x = array(transpose(bstar)[:n,:], dtype=result_t, copy=True)
if results['rank'] == n and m > n:
resids = array([np.linalg.norm(v[n:], 2)**2 for v in bstar])
st = s[:min(n, m)].copy().astype(_realType(result_t))
return wrap(x), wrap(resids), results['rank'], st
def lstsq(a, b, rcond=-1):
"""
Return the least-squares solution to an equation.
Solves the equation `a x = b` by computing a vector `x` that minimizes
the norm `|| b - a x ||`.
Parameters
----------
a : array_like, shape (M, N)
Input equation coefficients.
b : array_like, shape (M,) or (M, K)
Equation target values. If `b` is two-dimensional, the least
squares solution is calculated for each of the `K` target sets.
rcond : float, optional
Cutoff for ``small`` singular values of `a`.
Singular values smaller than `rcond` times the largest singular
value are considered zero.
Returns
-------
x : ndarray, shape(N,) or (N, K)
Least squares solution. The shape of `x` depends on the shape of
`b`.
residues : ndarray, shape(), (1,), or (K,)
Sums of residues; squared Euclidian norm for each column in
`b - a x`.
If the rank of `a` is < N or > M, this is an empty array.
If `b` is 1-dimensional, this is a (1,) shape array.
Otherwise the shape is (K,).
rank : integer
Rank of matrix `a`.
s : ndarray, shape(min(M,N),)
Singular values of `a`.
Raises
------
LinAlgError
If computation does not converge.
Notes
-----
If `b` is a matrix, then all array results returned as
matrices.
Examples
--------
Fit a line, ``y = mx + c``, through some noisy data-points:
>>> x = np.array([0, 1, 2, 3])
>>> y = np.array([-1, 0.2, 0.9, 2.1])
By examining the coefficients, we see that the line should have a
gradient of roughly 1 and cuts the y-axis at more-or-less -1.
We can rewrite the line equation as ``y = Ap``, where ``A = [[x 1]]``
and ``p = [[m], [c]]``. Now use `lstsq` to solve for `p`:
>>> A = np.vstack([x, np.ones(len(x))]).T
>>> A
array([[ 0., 1.],
[ 1., 1.],
[ 2., 1.],
[ 3., 1.]])
>>> m, c = np.linalg.lstsq(A, y)[0]
>>> print m, c
1.0 -0.95
Plot the data along with the fitted line:
>>> import matplotlib.pyplot as plt
>>> plt.plot(x, y, 'o', label='Original data', markersize=10)
>>> plt.plot(x, m*x + c, 'r', label='Fitted line')
>>> plt.legend()
>>> plt.show()
"""
import math
a, _ = _makearray(a)
b, wrap = _makearray(b)
is_1d = len(b.shape) == 1
if is_1d:
b = b[:, newaxis]
_assertRank2(a, b)
m = a.shape[0]
n = a.shape[1]
n_rhs = b.shape[1]
ldb = max(n, m)
if m != b.shape[0]:
raise LinAlgError, 'Incompatible dimensions'
t, result_t = _commonType(a, b)
real_t = _linalgRealType(t)
bstar = zeros((ldb, n_rhs), t)
bstar[:b.shape[0], :n_rhs] = b.copy()
a, bstar = _fastCopyAndTranspose(t, a, bstar)
s = zeros((min(m, n), ), real_t)
nlvl = max(0, int(math.log(float(min(m, n)) / 2.)) + 1)
iwork = zeros((3 * min(m, n) * nlvl + 11 * min(m, n), ), fortran_int)
if isComplexType(t):
lapack_routine = lapack_lite.zgelsd
lwork = 1
rwork = zeros((lwork, ), real_t)
work = zeros((lwork, ), t)
results = lapack_routine(m, n, n_rhs, a, m, bstar, ldb, s, rcond, 0,
work, -1, rwork, iwork, 0)
lwork = int(abs(work[0]))
rwork = zeros((lwork, ), real_t)
a_real = zeros((m, n), real_t)
bstar_real = zeros((
ldb,
n_rhs,
), real_t)
results = lapack_lite.dgelsd(m, n, n_rhs, a_real, m, bstar_real, ldb,
s, rcond, 0, rwork, -1, iwork, 0)
lrwork = int(rwork[0])
work = zeros((lwork, ), t)
rwork = zeros((lrwork, ), real_t)
results = lapack_routine(m, n, n_rhs, a, m, bstar, ldb, s, rcond, 0,
work, lwork, rwork, iwork, 0)
else:
lapack_routine = lapack_lite.dgelsd
lwork = 1
work = zeros((lwork, ), t)
results = lapack_routine(m, n, n_rhs, a, m, bstar, ldb, s, rcond, 0,
work, -1, iwork, 0)
lwork = int(work[0])
work = zeros((lwork, ), t)
results = lapack_routine(m, n, n_rhs, a, m, bstar, ldb, s, rcond, 0,
work, lwork, iwork, 0)
if results['info'] > 0:
raise LinAlgError, 'SVD did not converge in Linear Least Squares'
resids = array([], t)
if is_1d:
x = array(ravel(bstar)[:n], dtype=result_t, copy=True)
if results['rank'] == n and m > n:
resids = array([np.linalg.norm(ravel(bstar)[n:], 2)**2])
else:
x = array(transpose(bstar)[:n, :], dtype=result_t, copy=True)
if results['rank'] == n and m > n:
resids = array([np.linalg.norm(v[n:], 2)**2 for v in bstar])
st = s[:min(n, m)].copy().astype(_realType(result_t))
return wrap(x), wrap(resids), results['rank'], st