在networkx图中查找给定长度的所有路径/人行道 [英] Finding all paths/walks of given length in a networkx graph

问题描述

我正在使用networkx并试图找到图中所有长度为3的步行，特别是具有三个边的路径.我试图在networkx文档中找到有关算法的一些信息，但我只能在图中找到最短路径的算法.我能找到一条路径穿过特定节点的长度，例如，如果最短路径是14-> 15-> 16，那么路径穿过节点14-> 11-> 12-> 16?这是一个示例的图形图像:

I'm using networkx and trying to find all the walks with length 3 in the graph, specifically the paths with three edges. I tried to find some information about the algorithms in the networkx documentation but I could only find the algorithms for the shortest path in the graph. Can I find a length of a path trough specific nodes, for example a path trough nodes 14 -> 11 -> 12 -> 16 if the shortest path is 14 -> 15 -> 16? Here's an image of a graph for an example:

推荐答案

最简单的版本(下面是另一个版本，我认为是更快的版本):

Simplest version (another version is below which I think is faster):

def findPaths(G,u,n):
    if n==0:
        return [[u]]
    paths = [[u]+path for neighbor in G.neighbors(u) for path in findPaths(G,neighbor,n-1) if u not in path]
    return paths

这需要一个网络G和一个节点u和一个长度n.它从u的不包括u的邻居开始递归地找到长度为n-1的所有路径.然后，它将u粘贴在每个这样的路径的前面，并返回这些路径的列表.

This takes a network G and a node u and a length n. It recursively finds all paths of length n-1 starting from neighbors of u that don't include u. Then it sticks u at the front of each such path and returns a list of those paths.

注意，每个路径都是一个有序列表.它们都从指定的节点开始.因此，对于您想要的内容，只需在其周围包裹一个循环:

Note, each path is an ordered list. They all start from the specified node. So for what you want, just wrap a loop around this:

allpaths = []
for node in G:
    allpaths.extend(findPaths(G,node,3))

请注意，该路径将具有任何a-b-c-d路径以及反向的d-c-b-a路径.

Note that this will have any a-b-c-d path as well as the reverse d-c-b-a path.

如果您难以理解列表理解"，那么这里有一个等效的选择:

If you find the "list comprehension" to be a challenge to interpret, here's an equivalent option:

def findPathsNoLC(G,u,n):
    if n==0:
        return [[u]]
    paths = []
    for neighbor in G.neighbors(u):
        for path in findPathsNoLC(G,neighbor,n-1):
            if u not in path:
                paths.append([u]+path)
    return paths

为优化此效果，尤其是在有多个循环的情况下，可能值得发送一组不允许的节点.在每个嵌套调用中，它将知道不包含递归中更高层的任何节点.这将代替if u not in path检查工作.该代码将更难以理解，但运行速度更快.

For optimizing this, especially if there are many cycles, it may be worth sending in a set of disallowed nodes. At each nested call it would know not to include any nodes from higher up in the recursion. This would work instead of the if u not in path check. The code would be a bit more difficult to understand, but it would run faster.

def findPaths2(G,u,n,excludeSet = None):
    if excludeSet == None:
        excludeSet = set([u])
    else:
        excludeSet.add(u)
    if n==0:
        return [[u]]
    paths = [[u]+path for neighbor in G.neighbors(u) if neighbor not in excludeSet for path in findPaths2(G,neighbor,n-1,excludeSet)]
    excludeSet.remove(u)
    return paths

请注意，我必须在递归调用之前将u添加到excludeSet，然后在返回之前将其删除.

Note that I have to add u to excludeSet before the recursive call and then remove it before returning.

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