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numpy.linalg.tensorinv(a, ind=2)
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Compute the ?inverse? of an N-dimensional array.
The result is an inverse for
a
relative to the tensordot operationtensordot(a, b, ind)
, i. e., up to floating-point accuracy,tensordot(tensorinv(a), a, ind)
is the ?identity? tensor for the tensordot operation.Parameters: a : array_like
Tensor to ?invert?. Its shape must be ?square?, i. e.,
prod(a.shape[:ind]) == prod(a.shape[ind:])
.ind : int, optional
Number of first indices that are involved in the inverse sum. Must be a positive integer, default is 2.
Returns: b : ndarray
a
?s tensordot inverse, shapea.shape[ind:] + a.shape[:ind]
.Raises: LinAlgError
If
a
is singular or not ?square? (in the above sense).See also
tensordot
,tensorsolve
Examples
>>> a = np.eye(4*6) >>> a.shape = (4, 6, 8, 3) >>> ainv = np.linalg.tensorinv(a, ind=2) >>> ainv.shape (8, 3, 4, 6) >>> b = np.random.randn(4, 6) >>> np.allclose(np.tensordot(ainv, b), np.linalg.tensorsolve(a, b)) True
>>> a = np.eye(4*6) >>> a.shape = (24, 8, 3) >>> ainv = np.linalg.tensorinv(a, ind=1) >>> ainv.shape (8, 3, 24) >>> b = np.random.randn(24) >>> np.allclose(np.tensordot(ainv, b, 1), np.linalg.tensorsolve(a, b)) True
numpy.linalg.tensorinv()
2017-01-10 18:14:49
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