def tensorinv(a, ind=<NUM_LIT:2>):
a = asarray(a)<EOL>oldshape = a.shape<EOL>prod = <NUM_LIT:1><EOL>if ind > <NUM_LIT:0>:<EOL><INDENT>invshape = oldshape[ind:] + oldshape[:ind]<EOL>for k in oldshape[ind:]:<EOL><INDENT>prod *= k<EOL><DEDENT><DEDENT>else:<EOL><INDENT>raise ValueError("<STR_LIT>")<EOL><DEDENT>a = a.reshape(prod, -<NUM_LIT:1>)<EOL>ia = inv(a)<EOL>return ia.reshape(*invshape)<EOL>
Compute the 'inverse' of an N-dimensional array.

The result is an inverse for `a` relative to the tensordot operation
``tensordot(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, shape ``a.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