numpy.linalg.tensorsolve
-
numpy.linalg.tensorsolve(a, b, axes=None)
[source] -
Solve the tensor equation
a x = b
for x.It is assumed that all indices of
x
are summed over in the product, together with the rightmost indices ofa
, as is done in, for example,tensordot(a, x, axes=b.ndim)
.Parameters: a : array_like
Coefficient tensor, of shape
b.shape + Q
.Q
, a tuple, equals the shape of that sub-tensor ofa
consisting of the appropriate number of its rightmost indices, and must be such thatprod(Q) == prod(b.shape)
(in which sensea
is said to be ‘square’).b : array_like
Right-hand tensor, which can be of any shape.
axes : tuple of ints, optional
Axes in
a
to reorder to the right, before inversion. If None (default), no reordering is done.Returns: x : ndarray, shape Q
Raises: LinAlgError
If
a
is singular or not ‘square’ (in the above sense).See also
Examples
>>> a = np.eye(2*3*4) >>> a.shape = (2*3, 4, 2, 3, 4) >>> b = np.random.randn(2*3, 4) >>> x = np.linalg.tensorsolve(a, b) >>> x.shape (2, 3, 4) >>> np.allclose(np.tensordot(a, x, axes=3), b) True
© 2008–2017 NumPy Developers
Licensed under the NumPy License.
https://docs.scipy.org/doc/numpy-1.13.0/reference/generated/numpy.linalg.tensorsolve.html