def test_fullrank(self):
X = standard_normal((40,10))
X[:,0] = X[:,1] + X[:,2]
Y = tools.fullrank(X)
self.assertEquals(Y.shape, (40,9))
self.assertEquals(tools.rank(Y), 9)
X[:,5] = X[:,3] + X[:,4]
Y = tools.fullrank(X)
self.assertEquals(Y.shape, (40,8))
self.assertEquals(tools.rank(Y), 8)
def contrastfromcols(L, D, pseudo=None):
"""
From an n x p design matrix D and a matrix L, tries
to determine a p x q contrast matrix C which
determines a contrast of full rank, i.e. the
n x q matrix
dot(transpose(C), pinv(D))
is full rank.
L must satisfy either L.shape[0] == n or L.shape[1] == p.
If L.shape[0] == n, then L is thought of as representing
columns in the column space of D.
If L.shape[1] == p, then L is thought of as what is known
as a contrast matrix. In this case, this function returns an estimable
contrast corresponding to the dot(D, L.T)
Note that this always produces a meaningful contrast, not always
with the intended properties because q is always non-zero unless
L is identically 0. That is, it produces a contrast that spans
the column space of L (after projection onto the column space of D).
Parameters
----------
L : array-like
D : array-like
"""
L = np.asarray(L)
D = np.asarray(D)
n, p = D.shape
if L.shape[0] != n and L.shape[1] != p:
raise ValueError("shape of L and D mismatched")
if pseudo is None:
pseudo = np.linalg.pinv(D) # D^+ \approx= ((dot(D.T,D))^(-1),D.T)
if L.shape[0] == n:
C = np.dot(pseudo, L).T
else:
C = L
C = np.dot(pseudo, np.dot(D, C.T)).T
Lp = np.dot(D, C.T)
if len(Lp.shape) == 1:
Lp.shape = (n, 1)
if rank(Lp) != Lp.shape[1]:
Lp = fullrank(Lp)
C = np.dot(pseudo, Lp).T
return np.squeeze(C)
def contrastfromcols(L, D, pseudo=None):
"""
From an n x p design matrix D and a matrix L, tries
to determine a p x q contrast matrix C which
determines a contrast of full rank, i.e. the
n x q matrix
dot(transpose(C), pinv(D))
is full rank.
L must satisfy either L.shape[0] == n or L.shape[1] == p.
If L.shape[0] == n, then L is thought of as representing
columns in the column space of D.
If L.shape[1] == p, then L is thought of as what is known
as a contrast matrix. In this case, this function returns an estimable
contrast corresponding to the dot(D, L.T)
Note that this always produces a meaningful contrast, not always
with the intended properties because q is always non-zero unless
L is identically 0. That is, it produces a contrast that spans
the column space of L (after projection onto the column space of D).
Parameters
----------
L : array-like
D : array-like
"""
L = np.asarray(L)
D = np.asarray(D)
n, p = D.shape
if L.shape[0] != n and L.shape[1] != p:
raise ValueError("shape of L and D mismatched")
if pseudo is None:
pseudo = np.linalg.pinv(D) # D^+ \approx= ((dot(D.T,D))^(-1),D.T)
if L.shape[0] == n:
C = np.dot(pseudo, L).T
else:
C = L
C = np.dot(pseudo, np.dot(D, C.T)).T
Lp = np.dot(D, C.T)
if len(Lp.shape) == 1:
Lp.shape = (n, 1)
if rank(Lp) != Lp.shape[1]:
Lp = fullrank(Lp)
C = np.dot(pseudo, Lp).T
return np.squeeze(C)