Question

我正在Python 3中使用floyd_warshall_predecessor_and_distance的{{1}}函数在双向图上找到最短路径。该函数返回给定的两个节点（如果存在边）和路径的一部分之间的最短距离。我将通过说“一部分”来澄清我的意思。以下是我的输入和输出。

输入：

Networkx

输出：

import networkx as nx
import numpy as np
import matplotlib.pyplot as plt
np.random.seed(0)
V = [1, 2, 3, 4, 5]
N = [(i,j) for i in V for j in V if i!=j]
E = {} #Creating an empty dictionary to store the travel times from node i to j
Elist = (list(np.random.randint(low=1, high = 30, size = len(N))))
for i in range(len(N)):
    E[N[i]] = Elist[i] # (i,j) does not have to be equal to (j,i)
E[(2, 1)] = 5
E[(5, 4)] = 0
E[(2, 4)] = 20
G=nx.DiGraph()
G.add_nodes_from(V)
for i in E:
    G.add_weighted_edges_from([(i[0], i[1], E[i])])
path_lengths=nx.floyd_warshall_predecessor_and_distance(G, weight='weight')
path_lengths

我故意为（2，4）设置了2> 1> 5> 4的路径。当我查看({1: {2: 1, 3: 4, 4: 5, 5: 1}, 2: {1: 2, 3: 4, 4: 5, 5: 1}, 3: {1: 3, 2: 3, 4: 5, 5: 1}, 4: {1: 4, 2: 1, 3: 4, 5: 1}, 5: {1: 4, 2: 5, 3: 4, 4: 5}}, {1: defaultdict(<function networkx.algorithms.shortest_paths.dense.floyd_warshall_predecessor_and_distance.<locals>.<lambda>.<locals>.<lambda>()>, {1: 0, 2: 13, 3: 8, 4: 1, 5: 1}), 2: defaultdict(<function networkx.algorithms.shortest_paths.dense.floyd_warshall_predecessor_and_distance.<locals>.<lambda>.<locals>.<lambda>()>, {2: 0, 1: 5, 3: 13, 4: 6, 5: 6}), 3: defaultdict(<function networkx.algorithms.shortest_paths.dense.floyd_warshall_predecessor_and_distance.<locals>.<lambda>.<locals>.<lambda>()>, {3: 0, 1: 10, 2: 20, 4: 11, 5: 11}), 4: defaultdict(<function networkx.algorithms.shortest_paths.dense.floyd_warshall_predecessor_and_distance.<locals>.<lambda>.<locals>.<lambda>()>, {4: 0, 1: 5, 2: 18, 3: 7, 5: 6}), 5: defaultdict(<function networkx.algorithms.shortest_paths.dense.floyd_warshall_predecessor_and_distance.<locals>.<lambda>.<locals>.<lambda>()>, {5: 0, 1: 5, 2: 13, 3: 7, 4: 0})})时，我看到从节点2到4，我停了5个。此外，从2转到5，我停了1，这两个告诉我完整的路线，但是我希望将整个路线视为输出，例如path_lengths[0]或2: {... 4: 1, 5, ...}而不是看到它的一部分，然后将各个部分组合在一起。 Networkx中有没有更好的软件包可以做到这一点？顺便说一下，我的双向图不包含负权重，并且该图可能很大（这就是为什么我选择此函数的原因）。

这是我尝试开始的地方：

{(2,4): (2,1), (1,5), (5,4)}

感谢您的回复！

Answer 1

我找到了解决问题的办法。以下代码创建两个字典。对于paths，键表示圆弧，值表示最短路径需要采取的连续圆弧。对于shortest_distance，键表示圆弧，值表示最短的距离。我将其留在此处以供将来参考。

输入：

def arcs(seq, n):
    return [seq[max(i, 0):i + n] for i in range(-n + 1, len(seq))]
paths = {}; shortest_distance = {}
for v in V:
    for d in V:
        if v!=d:
            path = nx.single_source_dijkstra_path(G,v)
            paths[(v,d)] = path[d]
for i in paths:
    paths[i] = (arcs(paths[i],2)[1:-1])
    shortest_distance[(i[0],i[1])] = path_lengths[1][i[0]][i[1]]
    for j in range(len(paths[i])):
        paths[i][j] = tuple(paths[i][j])    
for i in paths:
    print(i, paths[i], shortest_distance[i])

输出：

(1, 2) [(1, 2)] 13
(1, 3) [(1, 5), (5, 4), (4, 3)] 8
(1, 4) [(1, 5), (5, 4)] 1
(1, 5) [(1, 5)] 1
(2, 1) [(2, 1)] 5
(2, 3) [(2, 1), (1, 5), (5, 4), (4, 3)] 13
(2, 4) [(2, 1), (1, 5), (5, 4)] 6
(2, 5) [(2, 1), (1, 5)] 6
(3, 1) [(3, 1)] 10
(3, 2) [(3, 2)] 20
(3, 4) [(3, 1), (1, 5), (5, 4)] 11
(3, 5) [(3, 1), (1, 5)] 11
(4, 1) [(4, 1)] 5
(4, 2) [(4, 1), (1, 2)] 18
(4, 3) [(4, 3)] 7
(4, 5) [(4, 1), (1, 5)] 6
(5, 1) [(5, 4), (4, 1)] 5
(5, 2) [(5, 2)] 13
(5, 3) [(5, 4), (4, 3)] 7
(5, 4) [(5, 4)] 0

如何从Networkx最短路径算法接收完整路径

1 个答案: