如何在 python 中将矩阵的特定行/列相乘？

Question

I have to input matrices of shape

m1: (n,3)
m2: (n,3)

I want to multiply each row (each n of size 3) with its correspondence of the other matrix, such that i get a (3,3) matrix for each row.

When im trying to just use eg m1[0]@m2.T[0] the operation doesnt work, as m[0] delivers a (3,) list instead of a (3,1) matrix, on which i could use matrix operations.

Is there a relatively easy or elegant way to get the desired (3,1) matrix for the matrix multiplication?

Answer 1

Generally, I would recommend using np.einsum for most matrix operations as it very elegant. To obtain a the row-wise outer product of the vectors contained in m1 and m2 of shape (n, 3) you could do the following:

import numpy as np
m1 = np.array([1, 2, 3]).reshape(1, 3)
m2 = np.array([1, 2, 3]).reshape(1, 3)
result = np.einsum("ni, nj -> nij", m1, m2)
print(result)
>>>array([[[1, 2, 3],
        [2, 4, 6],
        [3, 6, 9]]])

Answer 2

By default, numpy gets rid of the singleton dimension, as you have noticed.
You can use np.newaxis (or equivalently None . That is an implementation detail, but also works in pytorch) for the second axis to tell numpy to "invent" a new one.

import numpy as np
a = np.ones((3,3))
a[1].shape                 # this is (3,)
a[1,:].shape               # this is (3,)
a[1][...,np.newaxis].shape # this is (3,1)

However , you can also use dot or outer directly:

>>> a = np.eye(3)
>>> np.outer(a[1], a[1])
array([[0., 0., 0.],
       [0., 1., 0.],
       [0., 0., 0.]])
>>> np.dot(a[1], a[1])
1.0