numpy.linalg.tensorsolve — NumPy v1.15 Manual (original) (raw)

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 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 thatprod(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).

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 thatprod(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).

Examples

a = np.eye(234) a.shape = (23, 4, 2, 3, 4) b = np.random.randn(23, 4) x = np.linalg.tensorsolve(a, b) x.shape (2, 3, 4) np.allclose(np.tensordot(a, x, axes=3), b) True