计算两个 3D 点的 numpy 数组之间最小距离的快速方法

Question

我想知道是否有一种快速的方法来计算 3D numpy 数组 (A [N,3]) 的所有点与第二个 3D numpy 数组 (B [M,3]) 的所有点之间的欧几里得距离。

然后我应该得到一个数组C，这将是[N, M]，从数组A的点到数组B的点的所有距离，然后沿指定轴使用np.min()来获得所有最小距离集合点A 到集合点B。

这是我迄今为止完成的实现方式：

distances = np.repeat(9999, len(A))
for i, point in enumerate(A):
  min_distance = np.min(np.sqrt(np.sum(np.square(point - B), axis=1)))
  distances[i] = min_distance

有什么办法可以摆脱 for 循环...？

提前致谢:)

Answer 1

如果scipy方法不起作用或者你确实有其他原因，这里有一个numpy方法-

import numpy as np


x = np.random.random((200, 3))
y = np.random.random((100,3))

x = x.reshape((-1, 1, 3))                 # [200x1x3]
y = np.expand_dims(y, axis=0)             # [1x100x3]
y = y.repeat(x.shape[0], axis=0)          # [200x100x3]

distance = np.linalg.norm(y-x, axis=2)    # Difference is [200x100x3] and norm results in [200x100]

Answer 2

import numpy as np

# arrays with xyz coordinates of all points 
a = np.asarray([[xa1,ya1,za1],...,[xan,yan,zan]])
b = np.asarray([[xb1,yb1,zb1],...,[xbn,ybn,zbn]])

# reshaping to be able to calculate the distance matrix
a_reshaped = a.reshape(a.shape[0], 1, 3)
b_reshaped = b.reshape(1, b.shape[0], 3)

"""calculation of all distances between all points  - creates a 
len(a) x len(b) matrix""" 
distance = np.sqrt(np.sum((a_reshaped - b_reshaped)**2, axis=2))