python中的张量点操作 [英] tensor dot operation in python

问题描述

我有两个数组A=[1,2,3]和B=[[1],[0],[1],[0]].问题是如何在python中执行其张量点积.我期望得到:

I have two arrays A=[1,2,3] and B=[[1],[0],[1],[0]]. The question how to perform their tensor dot product in python. I am expecting to get:

C=[[1,2,3],
   [0,0,0],
   [1,2,3],
   [0,0,0]]

np.tensordot()函数返回有关数组形状的错误.

The function np.tensordot() returns an error concerning shapes of arrays.

这个问题的补充.如果矩阵的形状完全不同，该怎么做?

A little addition to this question. How to do such operation if matrix are totally different in shape, like:

A=[[1,1,1,1],
   [1,1,1,1],
   [2,2,2,2],
   [3,3,3,3]]

B=[2,1]

C=[[[2,1],[2,1],[2,1],[2,1]],
   [[2,1],[2,1],[2,1],[2,1]],
   [[4,2],[4,2],[4,2],[4,2]],
   [[6,3],[6,3],[6,3],[6,3]]]

推荐答案

尝试使用正确的numpy数组:

>>> array([[1],[2],[3]]).dot(array([[1,0,1,0]]))
array([[1, 0, 1, 0],
       [2, 0, 2, 0],
       [3, 0, 3, 0]])

如果对齐方式不同，则使用a.transpose()可以将其翻转:

If your alignment is different, using a.transpose() can flip it:

>>> array([[1],[2],[3]]).dot(array([[1,0,1,0]])).transpose()
array([[1, 2, 3],
       [0, 0, 0],
       [1, 2, 3],
       [0, 0, 0]])

如果(出于某种原因)必须使用tensordot()，请尝试以下操作:

If you (for whatever reason) have to use tensordot(), try this:

>>> numpy.tensordot([1,2,3], [1,0,1,0], axes=0)
array([[1, 0, 1, 0],
       [2, 0, 2, 0],
       [3, 0, 3, 0]])

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