在 Python 中找到坐標系中某些點之間的最短路徑

Question

我編寫了一個代碼，可以在坐標系中的某個寬度和長度范圍內產生所需數量的點。 它計算我使用歐幾里得方法產生的這些點的距離矩陣並將其制成表格。

我的代碼在這里：

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)

Output 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

我想創建一種算法，從輸入的隨機點開始，圍繞最短路徑中的所有點。 （最近鄰法繼續根據歐幾里得距離找到離起點最近的點，然后去未糾纏的點中離這個新點最近的點。這個過程一直持續到遍歷完所有點，完成一輪）。 我怎樣才能在 10 個不同的點重復這個過程 10 次，並得到一個像這樣的 output：

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
.
...

非常感謝您的幫助。

Answer 1

如果我正確理解了您的問題，那么這應該可以完成單條路徑的工作。

import random
import pandas as pd
from scipy.spatial import distance_matrix, 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=', ')

mat_dist = distance.cdist(sample, sample, 'euclidean')
df_mat_dist = pd.DataFrame(mat_dist)
print(df_mat_dist)

#Randomly select the first point
closest_idx = random.randrange(npoints)
path_points = [closest_idx]

#Find the closest point to the starting point, different from diagonal and save results
path_length = 0

for _ in range(npoints-1):
    closest_dist = df_mat_dist.loc[closest_idx, ~df_mat_dist.index.isin(path_points)].min()
    closest_idx = df_mat_dist.loc[closest_idx, ~df_mat_dist.index.isin(path_points)].idxmin()
    path_points.append(closest_idx)
    path_length += closest_dist

print(path_points, path_length)

Output

Type the npoints: 5
Enter the Width you want: 6
Enter the Height you want: 7
(2.45, 6.66), (3.01, 3.94), (5.06, 0.51), (5.89, 1.04), (1.37, 5.03)
          0         1         2         3         4
0  0.000000  2.775327  6.677550  6.587089  1.950042
1  2.775327  0.000000  3.993631  4.086550  1.970787
2  6.677550  3.993631  0.000000  0.988898  5.834766
3  6.587089  4.086550  0.988898  0.000000  6.030719
4  1.950042  1.970787  5.834766  6.030719  0.000000
[1, 4, 0, 3, 2] 11.49681560383563

從那你應該能夠調整代碼運行 10 次。