NumPy广播:计算两个数组之间的平方差之和

本文介绍了NumPy广播:计算两个数组之间的平方差之和的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有以下代码.它在Python中永远存在.必须有一种方法可以将该计算结果转换为广播...

I have the following code. It is taking forever in Python. There must be a way to translate this calculation into a broadcast...

def euclidean_square(a,b): squares = np.zeros((a.shape[0],b.shape[0])) for i in range(squares.shape[0]): for j in range(squares.shape[1]): diff = a[i,:] - b[j,:] sqr = diff**2.0 squares[i,j] = np.sum(sqr) return squares

推荐答案

您可以使用 np.einsum 在计算 broadcasted way ，就像这样-

You can use np.einsum after calculating the differences in a broadcasted way, like so -

ab = a[:,None,:] - b out = np.einsum('ijk,ijk->ij',ab,ab)

或使用 ，其可选的度量参数设置为'sqeuclidean'，以根据问题的需要为我们提供平方的欧几里德距离，就像这样-

Or use scipy's cdist with its optional metric argument set as 'sqeuclidean' to give us the squared euclidean distances as needed for our problem, like so -

from scipy.spatial.distance import cdist out = cdist(a,b,'sqeuclidean')