识别点具有最小欧几里得距离 [英] Identifying points with the smallest Euclidean distance

问题描述

我有n个三维点的集合，我想要找到其中2个是最接近。我可以拿出2的尺寸最好是：

I have a collection of n dimensional points and I want to find which 2 are the closest. The best I could come up for 2 dimensions is:

from numpy import *
myArr = array( [[1, 2],
                [3, 4],
                [5, 6],
                [7, 8]] )

n = myArr.shape[0]
cross = [[sum( ( myArr[i] - myArr[j] ) ** 2 ), i, j]
         for i in xrange( n )
         for j in xrange( n )
         if i != j
         ]

print min( cross )

这使得

[8, 0, 1]

但是，这是对于大型阵列太慢。我能将它什么样的优化？

But this is too slow for large arrays. What kind of optimisation can I apply to it?

相关报道：

欧氏距离，而不是在

推荐答案

尝试 scipy.spatial.distance.pdist（myArr，该）。这会给你一个浓缩的距离矩阵。您可以使用 argmin 就可以了，找到最小的值的索引。这可以转换成一对信息。

Try scipy.spatial.distance.pdist(myArr). This will give you a condensed distance matrix. You can use argmin on it and find the index of the smallest value. This can be converted into the pair information.

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