我编写了一个代码,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。它计算并列出我使用欧几里德方法生成的这些点的距离矩阵。
我的代码在这里:
import pandas as pd
from scipy.spatial import distance_matrix, distance
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=', ')
mat_dist = distance.cdist(sample, sample, 'euclidean')
df_mat_dist = pd.DataFrame(mat_dist)
print(df_mat_dist)
输出是:
Type the npoints:5
Enter the Width you want:6
Enter the Height you want:7
(3.25, 3.55), (5.51, 6.47), (5.87, 5.31), (2.27, 3.20), (0.96, 3.83)
0 1 2 3 4
0 0.000000 3.690201 3.153510 1.047022 2.305800
1 3.690201 0.000000 1.209096 4.608588 5.257688
2 3.153510 1.209096 0.000000 4.176733 5.123103
3 1.047022 4.608588 4.176733 0.000000 1.450613
4 2.305800 5.257688 5.123103 1.450613 0.000000
Process finished with exit code 0
我想创建一种算法,从输入的随机点开始,绕过最短路径中的所有点。 (最近邻法继续根据欧几里得距离找到离起始点最近的点,然后去未纠缠点中距离这个新点最近的点。这个过程一直持续到遍历完所有点,完成一轮)。我怎样才能在 10 个不同的点重复这个过程 10 次并得到如下输出:
Tour Number:1
Number of points visited in order in the relevant round: 0-7-3-8-2...
Total route length of the tour: 18,75755
Tour Number:2
The number of the points visited in order in the relevant round: 6-9-11-2-7...
Total route length of the tour: 14,49849
.
...
非常感谢您的帮助。