用 Python 中的随机点制作欧几里得距离矩阵

Question

我编写了一段代码，可以在坐标系中的特定宽度和长度范围内生成所需数量的点。我如何计算和制表我使用欧几里得方法生成的这些点的距离矩阵？

import random

npoints = int(input("Type the npoints:"))
width = float(input("Enter the Width you want:"))
height = float(input("Enter the Height you want:"))

sample = []
for _ in range(npoints):
    sample.append((width * random.random(), height * random.random()))
print(*[f"({w:.2f}, {h:.2f})" for w, h in sample], sep=', ')

输出为：

Type the npoints:4
Enter the Width you want:10
Enter the Height you want:10
(8.52, 3.73), (9.69, 6.87), (8.20, 6.14), (4.18, 0.76)

Process finished with exit code 0

我怎样才能像这个例子那样创建一个带有随机点的距离矩阵：

非常感谢您的帮助。

Answer 1

如果要使用外部模块，scipy矩阵计算效率很高

import random
import pandas as pd
from scipy.spatial import distance

npoints = int(input("Type the npoints:"))
width = float(input("Enter the Width you want:"))
height = float(input("Enter the Height you want:"))

sample = []
for _ in range(npoints):
    sample.append((width * random.random(), height * random.random()))
print(*[f"({w:.2f}, {h:.2f})" for w, h in sample], sep=', ')

#Create a matrix from these points
mat_dist = distance.cdist(sample, sample, 'euclidean')
df_mat_dist = pd.DataFrame(mat_dist)
print(df_mat_dist)

输出

Type the npoints:4
Enter the Width you want:10
Enter the Height you want:10
(8.89, 8.85), (9.00, 9.43), (9.67, 9.45), (3.96, 5.68)
          0         1         2         3
0  0.000000  0.584322  0.985072  5.856736
1  0.584322  0.000000  0.669935  6.277323
2  0.985072  0.669935  0.000000  6.839240
3  5.856736  6.277323  6.839240  0.000000

用 Python 中的随机点制作欧几里得距离矩阵

Making an Euclidean Distance Matrix with Random Points in Python

python

arrays

for-loop

matrix

euclidean-distance

输出