使用bfs和dfs networkx将字典的输出转换为列表

本文介绍了使用bfs和dfs networkx将字典的输出转换为列表的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我目前正在将networkx库用于带有BFS和DFS的Python。我需要得到一棵树，然后探索它以获取从起始节点到结束节点的路径。

I am currently using networkx library for Python with BFS and DFS. I need to get a tree and then explore it to get a path from a start node to an end node.

对于BFS部分，我正在使用 bfs_successors ，它从源返回广度优先搜索中的后继迭代器。

For the BFS part I am using bfs_successorsand it returns an iterator of successors in breadth-first-search from source.

对于DFS部分，我正在使用： dfs_successors ，它从源返回深度优先搜索的后继字典。

For the DFS part I am using: dfs_successors and it returns a dictionary of successors in depth-first-search from source.

我需要从两种算法中获得从源到末端的节点列表。每个节点是（x，y），并且是网格中的一个单元。

I need to get a list of nodes from source to end from both the algorithms. Each node is (x, y) and is a cell in a grid.

您对此有任何建议吗？

Do you have any advice about how to do it? Can you help me please?

MWE：

DFS = nx.bfs_successors(mazePRIM,start) print(dict(BFS)) DFS = nx.dfs_successors(mazePRIM, start) print(DFS)

我明白了：

{(0, 0): [(0, 1), (1, 0)], (1, 0): [(1, 1)], (1, 1): [(1, 2)], (1, 2): [(0, 2), (1, 3)], (0, 2): [(0, 3)]} {(0, 0): [(0, 1), (1, 0)], (1, 0): [(1, 1)], (1, 1): [(1, 2)], (1, 2): [(0, 2), (1, 3)], (0, 2): [(0, 3)]}

但是我需要这样的输出：

But I need an output like this:

[(0, 0), (1, 0), (1, 1), (1, 2), (1, 3)]

从头到尾的节点列表。

推荐答案

IIUC，您对找到所有受 nx.bfs_successors ，因为您只需要源节点和目标节点之间的路径。

IIUC you're not really interested in finding all successors encourtered with nx.bfs_successors, since you only need the path between a source and a target nodes.

为此，您可以找到最短路径（如果有多个路径）：

For that you can either find the shortest path (in the case there are multiple):

nx.shortest_path(G, source, target)

或查找它们之间的所有简单路径：

Or find all simple paths between them:

nx.all_simple_paths(G, source, target)

这将返回一个生成器，其中包含两个节点之间的所有简单路径。

Which returns a generator with all simple paths between both nodes.