Question

从API我得到一个包含不同坐标的“列表列表”：

List = [[1.0, 2.5, 3.6], [2.02, 2.3, 3.1], [1.5, 6.5, 3.9]]

我必须找到两个坐标之间的最小距离。我做了类似的事情：

MinDist = 9999999999.
for Coord1 in List:
    for Coord2 in List:
        if Coord1 != Coord2:
            Dist = CalcDistance(Coord1,Coord2)
            if Dist < MinDist:
                MinDist=Dist

是否有更“智能”（且更快）的方式来获取此信息？

Answer 1

假设CalcDistance类似于以下内容，您可以使用min和一个关键功能以及itertools.combinations

from itertools import zip_longest, combinations

def CalcDistance(a, b):
    return (sum((x-y)**2 for x, y in zip_longest(a, b, fillvalue=0)))**.5

List = [[1.0, 2.5, 3.6], [2.02, 2.3, 3.1], [1.5, 6.5, 3.9]]

print(min(combinations(List, 2), key=lambda x: CalcDistance(*x)))
# ([1.0, 2.5, 3.6], [2.02, 2.3, 3.1])

Answer 2

为什么不使用内置算法：

import numpy as np
from scipy.spatial.distance import pdist, squareform

List = [[1.0, 2.5, 3.6], [2.02, 2.3, 3.1], [1.5, 6.5, 3.9]]

dist_mat = squareform(pdist(List, CalcDistance))
np.fill_diagonal(dist_mat,np.inf)
i,j = np.unravel_index(dist_mat.argmin(), dist_mat.shape)

print((List[i],List[j]))

上面的代码结合了Find the index of the min value in a pdist condensed distance matrix和Numpy minimum in (row, column) format

pythonic方法比较循环列表的值

2 个答案: