Module « numpy.linalg »
Signature de la fonction tensorsolve
def tensorsolve(a, b, axes=None)
Description
tensorsolve.__doc__
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 of `a`, 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 of `a` consisting of the appropriate
number of its rightmost indices, and must be such that
``prod(Q) == prod(b.shape)`` (in which sense `a` 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
--------
numpy.tensordot, tensorinv, numpy.einsum
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
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