numpy.linalg.tensorinv

linalg.tensorinv(a, ind=2)

计算 N 维数组的“逆”.

结果是相对于张量点积运算 tensordot(a, b, ind) 的 a 的逆,即,直到浮点精度, tensordot(tensorinv(a), a, ind) 是张量点积运算的“恒等”张量.

参数:

aarray_like

要“反转”的张量. 它的形状必须是“正方形”,即 prod(a.shape[:ind]) == prod(a.shape[ind:]) .

ind整数,可选

参与逆和的第一个索引的数量.必须是正整数,默认为 2.

返回:

bndarray

a 的 tensordot 逆,形状为 a.shape[ind:] + a.shape[:ind] .

提出:

LinAlgError

如果 a 是奇异的或不是“方阵”(在上述意义上).

参见

numpy.tensordot , tensorsolve

示例

Try it in your browser!

>>> import numpy as np
>>> a = np.eye(4*6)
>>> a.shape = (4, 6, 8, 3)
>>> ainv = np.linalg.tensorinv(a, ind=2)
>>> ainv.shape
(8, 3, 4, 6)
>>> rng = np.random.default_rng()
>>> b = rng.normal(size=(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)
>>> rng = np.random.default_rng()
>>> b = rng.normal(size=24)
>>> np.allclose(np.tensordot(ainv, b, 1), np.linalg.tensorsolve(a, b))
True

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