N个点，并在numpy的/ SciPy的基准之间的有效距离计算 [英] Efficient distance calculation between N points and a reference in numpy/scipy

问题描述

我刚开始使用SciPy的/ numpy的。我有一个100000 * 3阵列，每一行是一个坐标，和一个1 * 3的中心点。欲计算每一行的距离阵列到中心中，并将它们存储在另一个阵列。什么是最有效的方式做到这一点？

I just started using scipy/numpy. I have an 100000*3 array, each row is a coordinate, and a 1*3 center point. I want to calculate the distance for each row in the array to the center and store them in another array. What is the most efficient way to do it?

推荐答案

我要看看 scipy.spatial.distance.cdist ：

<一个href=\"http://docs.scipy.org/doc/scipy/reference/generated/scipy.spatial.distance.cdist.html\">http://docs.scipy.org/doc/scipy/reference/generated/scipy.spatial.distance.cdist.html

import numpy as np
import scipy

a = np.random.normal(size=(10,3))
b = np.random.normal(size=(1,3))

dist = scipy.spatial.distance.cdist(a,b) # pick the appropriate distance metric 

DIST 默认遥远的指标等同于：

dist for the default distant metric is equivalent to:

np.sqrt(np.sum((a-b)**2,axis=1))  

虽然 cdist 是大数组更有效的（我的机器上为你的尺寸问题， cdist 更快通过

although cdist is much more efficient for large arrays (on my machine for your size problem, cdist is faster by a factor of ~35x).

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