Question

我有4个长度为160000的列表，分别为s，x，y，z。我列出了x，y，z的3d数组的列表（点）。我需要找到标准的所有点组合之间的距离，并使点的索引与列表s的索引匹配，以便获得满足要求的2个点的s值。我正在使用下面的代码。有没有更快的方法可以做到这一点？

import numpy as np

points = []
for i in range(len(xnew)):
    a = np.array((xnew[i],ynew[i],znew[i]))
    points.append(a)
for i in range(len(points)):
    for j in range(len(points)):
        d = np.sqrt(np.sum((points[i] - points[j]) ** 2))
        if d <= 4 and d >=3:
            print(s[i],s[j],d)

Answer 1

要使用想法 cdist和 np.where使处理向量化

代码

import numpy as np
import scipy.spatial.distance

# Distance between all pairs of points
d = scipy.spatial.distance.cdist(points, points)
# Pairs within threshold
indexes = np.where(np.logical_and(d>=3, d<=4))

for i, j in indexes:
    if i < j: # since distance is symmetric, not reporting j, i
      print(s[i],s[j],d[i][j])

如果d Matrix太大而无法容纳到内存中，请找到每个点到所有其他点的距离

for i in range(len(points)):
    # Distance from point i to all other points
    d = scipy.spatial.distance.cdist(points,[points[i]])
    # Points within threshold
    indexes = np.where(np.logical_and(d>=3, d<=4))
    
    for ind in indexes:
      if ind.size > 0:
        for j in ind:
          if i < j:   # since distance is symmetric, not reporting j, i
            print(s[i], s[j], d[j][0])

测试

points = [
  [1, 2, 3],
  [1.1, 2.2, 3.3],
  [4, 5, 6],
  [2, 3, 4]
]
s = [0, 1, 2, 3]

输出（两种方法）

2 3 3.4641016151377544

Answer 2

points = np.array([x, y, z]).T                         

t1, t2 = np.triu_indices(len(points), k= 1)                  # triangular indices

p1 = points[t1]
p2 = points[t2]

d = p1 - p2                       # displacements from p1 to p2
d = np.linalg.norm(d, axis= -1)   # distances     from p1 to p2

mask = (3 <= d) & (d <= 4)
indx = np.where(mask)                     # indices where distance is between 4 and 3

ans = np.array([ s[t1[i]], s[t2[i]], d[i] ]).T

试运行：

n = 10

x = np.random.randint(10, size= [n])          # dummy data
y = np.random.randint(10, size= [n])
z = np.random.randint(10, size= [n])

s = np.random.randint(10, size= [n])

运行上述代码后

points

>>> array([
       [9, 3, 5],
       [7, 8, 1],
       [0, 0, 2],
       [6, 7, 2],
       [4, 4, 3],
       [8, 0, 9],
       [5, 2, 6],
       [0, 8, 9],
       [2, 6, 9],
       [4, 8, 4]])

s

>>> array([4, 2, 9, 9, 8, 2, 7, 6, 0, 5])

for e in ans:
   print(*e)

>>> 9.0  8.0  3.7416573867739413
    9.0  5.0  3.0
    8.0  7.0  3.7416573867739413

计算两个3D点之间距离的更快方法

2 个答案: