将二维数组中的每一列与另一个二维数组中的每一列相乘 [英] Multiply each column from 2D array with each column from another 2D array

问题描述

我有两个 Numpy 数组 x，形状为 (m, i) 和 y，形状为 (m, j) (所以行数是一样的).我想将 x 的每一列与 y 的每一列元素相乘，以便结果的形状为 (m, i*j).

I have two Numpy arrays x with shape (m, i) and y with shape (m, j) (so the number of rows is the same). I would like to multiply each column of x with each column of y element-wise so that the result is of shape (m, i*j).

示例:

import numpy as np

np.random.seed(1)
x = np.random.randint(0, 2, (10, 3))
y = np.random.randint(0, 2, (10, 2))

这将创建以下两个数组 x:

This creates the following two arrays x:

array([[1, 1, 0],
       [0, 1, 1],
       [1, 1, 1],
       [0, 0, 1],
       [0, 1, 1],
       [0, 0, 1],
       [0, 0, 0],
       [1, 0, 0],
       [1, 0, 0],
       [0, 1, 0]])

和 y:

array([[0, 0],
       [1, 1],
       [1, 1],
       [1, 0],
       [0, 0],
       [1, 1],
       [1, 1],
       [1, 1],
       [0, 1],
       [1, 0]])

现在结果应该是:

array([[0, 0, 0, 0, 0, 0],
       [0, 0, 1, 1, 1, 1],
       [1, 1, 1, 1, 1, 1],
       [0, 0, 0, 0, 1, 0],
       [0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 1, 1],
       [0, 0, 0, 0, 0, 0],
       [1, 1, 0, 0, 0, 0],
       [0, 1, 0, 0, 0, 0],
       [0, 0, 1, 0, 0, 0]])

目前，我已经通过 x 和 y 列上的两个嵌套循环实现了这个操作:

Currently, I've implemented this operation with two nested loops over the columns of x and y:

def _mult(x, y):
    r = []
    for xc in x.T:
        for yc in y.T:
            r.append(xc * yc)
    return np.array(r).T

但是，我很确定一定有我似乎想不出的更优雅的解决方案.

However, I'm pretty sure that there must be a more elegant solution that I can't seem to come up with.

推荐答案

使用 NumPy 广播 -

(y[:,None]*x[...,None]).reshape(x.shape[0],-1)

说明

作为输入，我们有 -

y : 10 x 2
x : 10 x 3

使用 y[:,None]，我们在现有的两个暗度之间引入了一个新轴，从而创建了它的 3D 阵列版本.这将保持第一个轴作为 3D 版本中的第一个轴，并将第二个轴作为第三个轴推出.

With y[:,None], we are introducing a new axis between the existing two dims, thus creating a 3D array version of it. This keeps the first axis as the first one in 3D version and pushes out the second axis as the third one.

使用 x[...,None]，我们引入一个新轴作为最后一个轴，方法是将两个现有的暗淡作为前两个暗淡，从而产生 3D 数组版本.

With x[...,None], we are introducing a new axis as the last one by pushing up the two existing dims as the first two dims to result in a 3D array version.

总而言之，随着新轴的引入，我们有 -

To summarize, with the introduction of new axes, we have -

y : 10 x 1 x 2
x : 10 x 3 x 1

使用 y[:,None]*x[...,None]，y 和 都会有broadcastingcode>x，生成形状为 (10,3,2) 的输出数组.要获得形状 (10,6) 的最终输出数组，我们只需要将最后两个轴与该重塑合并.

With y[:,None]*x[...,None], there would be broadcasting for both y and x, resulting in an output array with a shape of (10,3,2). To get to the final output array of shape (10,6), we just need to merge the last two axes with that reshape.

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