我有一个由几个点的 x,y,z 坐标组成的矩阵.我想找到极值点，即相距最远的两点.
I have a matrix consisting of x,y,z coordinates of several points. I would like to find the extremum points i.e. the two points that are farthest apart.
我可以在 matlab 中找到一种方法，但我需要在 Python 中使用它
I could figure out a way in matlab, but i need it in Python
这是matlab中的代码
Here is the code in matlab
A = randint(500,3,[-5 5]); D=pdist(A); D=squareform(D); [N,I]=max(D(:)); [I_row, I_col] = ind2sub(size(D),I);pdist 给出点对之间的距离(i，j).squareform 给出矩阵输出在最后两个步骤中，我尝试找到矩阵 I_row、I_col 的索引..
pdist gives the distance between pairs of points(i,j). squareform gives the matrix output In last two steps I attempt to find the indices of the matrix I_row, I_col..
点 I_row 和 I_col 具有最大距离..
Points I_row and I_col have the max distance..
谁能建议我使用 Python 的有效方法，因为我所有的其他代码都使用 Python.
Could anybody suggest me an efficient way in python as all my other codes are in Python.
推荐答案如果你有 scipy，你就有大部分 matlab 核心函数的完全等效的:
If you have scipy, you have exact equivalent for most of matlab core functions :
from numpy import random, nanmax, argmax, unravel_index from scipy.spatial.distance import pdist, squareform A = random.randint(-5,5, (500,3)) D = pdist(A) D = squareform(D); N, [I_row, I_col] = nanmax(D), unravel_index( argmax(D), D.shape )您也可以使用 itertools 在纯 python 中获取它:
You can also get it in pure python using itertools :
from itertools import combinations from random import randint A = [[randint(-5,5) for coord in range(3)] for point in range(500)] def square_distance(x,y): return sum([(xi-yi)**2 for xi, yi in zip(x,y)]) max_square_distance = 0 for pair in combinations(A,2): if square_distance(*pair) > max_square_distance: max_square_distance = square_distance(*pair) max_pair = pair