将一个数组的每个元素乘以另一个数组的每个元素 [英] Multiplying every element of one array by every element of another array

问题描述

说我有两个数组，

import numpy as np


x = np.array([1, 2, 3, 4])
y = np.array([5, 6, 7, 8])

什么是最快，最Pythonic等的方式来获得一个新数组 z ，其元素数量等于 x.size * y.size ，其中元素是来自两个输入数组的每对元素(x_i，y_j)的乘积.

What's the fastest, most Pythonic, etc., etc. way to get a new array, z, with a number of elements equal to x.size * y.size, in which the elements are the products of every pair of elements (x_i, y_j) from the two input arrays.

换句话说，我正在寻找一个 z 数组，其中 z [k] 是 x [i] * y [j] .

To rephrase, I'm looking for an array z in which z[k] is x[i] * y[j].

一种简单但效率低下的方法如下:

A simple but inefficient way to get this is as follows:

z = np.empty(x.size * y.size)
counter = 0
for i in x:
    for j in y:
        z[counter] = i * j
        counter += 1

运行上面的代码表明，在此示例中， z 是

Running the above code shows that z in this example is

In [3]: z
Out[3]: 
array([  5.,   6.,   7.,   8.,  10.,  12.,  14.,  16.,  15.,  18.,  21.,
        24.,  20.,  24.,  28.,  32.])

推荐答案

在这里可以建议另外两种方法.

Two more approaches could be suggested here.

使用 矩阵乘以np.dot :

np.dot(x[:,None],y[None]).ravel()

使用 np.einsum :

np.einsum('i,j->ij',x,y).ravel()

运行时测试

In [31]: N = 10000
    ...: x = np.random.rand(N)
    ...: y = np.random.rand(N)
    ...: 

In [32]: %timeit np.dot(x[:,None],y[None]).ravel()
1 loops, best of 3: 302 ms per loop

In [33]: %timeit np.einsum('i,j->ij',x,y).ravel()
1 loops, best of 3: 274 ms per loop

与 @BilalAkil的答案 相同，但具有 ravel()>代替 flatten()作为更快的替代方法-

Same as @BilalAkil's answer but with ravel() instead of flatten() as a faster alternative -

In [34]: %timeit np.multiply.outer(x, y).ravel() 
1 loops, best of 3: 211 ms per loop

@BilalAkil的答案 :

In [35]: %timeit np.multiply.outer(x, y).flatten()
1 loops, best of 3: 451 ms per loop

@Tim Leathart的答案 :

In [36]: %timeit np.array([y * a for a in x]).flatten()
1 loops, best of 3: 766 ms per loop

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