networkx中图形的遍历级别顺序 [英] Traversing level order for the graph in networkx

问题描述

我试图将DiGraph转换为n元树，并以级别顺序或BFS显示节点.我的树与此相似，但更大，为了简单起见，使用以下示例:

I am trying to convert a DiGraph into n-ary tree and displaying the nodes in level order or BFS. My tree is similar to this, but much larger, for simplicity using this example:

G = networkx.DiGraph()
G.add_edges_from([('n', 'n1'), ('n', 'n2'), ('n', 'n3')])
G.add_edges_from([('n4', 'n41'), ('n1', 'n11'), ('n1', 'n12'), ('n1', 'n13')])
G.add_edges_from([('n2', 'n21'), ('n2', 'n22')])
G.add_edges_from([('n13', 'n131'), ('n22', 'n221')])

树:从以下问题借来的数据:

Tree: borrowed the data from this question:

n---->n1--->n11
 |     |--->n12
 |     |--->n13
 |           |--->n131 
 |--->n2              
 |     |---->n21     
 |     |---->n22     
 |            |--->n221 
 |--->n3

我正在为此目的使用networkx.DiGraph并成功创建了图形.这是我创建DiGraph的代码:

I am using networkx.DiGraph for this purpose and created the graph successfully. Here is my code for creating a DiGraph:

G = nx.DiGraph()
roots = set()
for l in raw.splitlines():
    if len(l):
        target, prereq = regex1.split(l)
        deps = tuple(regex2.split(prereq))
        print("Add node:") + target
        roots.add(target)
        G.add_node(target)
        for d in deps:
            if d:
                G.add_edge(target, d)

我正在从以下格式的大约200行的文件中读取所有数据，并尝试获取依赖关系树.我的图大约有100个节点，有600条边.

I am reading the all the data from a file with about 200 lines in the following format and trying to get a dependency tree. My graph is around 100 nodes with 600 edges.

AAA: BBB,CCC,DDD,
BBB:
DDD: EEE,FFF,GGG,KKK
GGG: AAA,BBB,III,LLL
....
...
..
.

在线查看了networkx文档之后，现在我可以使用以下代码对依赖项树进行拓扑排序，从而获得级别顺序输出.

After looking into the networkx docs online, now I can achieve the the level order output doing a topological sort on the dependency tree, with the below code.

order =  nx.topological_sort(G)
print "topological sort"
print order 

输出:

['n2', 'n3', 'n1', 'n21', 'n22', 'n11', 'n13', 'n12', 'n221', 'n131']

顺序似乎是正确的，但是由于我需要分批处理作业(这样可以节省时间)，而不是按顺序处理，因此我希望按级别排序的输出批次或使用BFS进行输出.实现此目标的最佳方法是什么?
例如:level [0:n]，例如:

The order seems correct, but since I need to process the jobs in a batch (which saves time) and not sequentially, I want the output in level ordered output batches or using BFS. What is the best way to achieve this ?
ex: level[0:n], ex:

0. ['n'] 
1. ['n2', 'n3', 'n1',] 
2. ['n21', 'n22', 'n11',] 
3. ['n13', 'n12', 'n221', 'n131'] 

推荐答案

您可以使用bfs_edges()函数以广度优先搜索的顺序获取节点列表.

You could use the bfs_edges() function to get a list of nodes in a breadth-first-search order.

In [1]: import networkx

In [2]: G = networkx.DiGraph()

In [3]: G.add_edges_from([('n', 'n1'), ('n', 'n2'), ('n', 'n3')])

In [4]: G.add_edges_from([('n4', 'n41'), ('n1', 'n11'), ('n1', 'n12'), ('n1', 'n13')])

In [5]: G.add_edges_from([('n2', 'n21'), ('n2', 'n22')])

In [6]: G.add_edges_from([('n13', 'n131'), ('n22', 'n221')])

In [7]: list(networkx.bfs_edges(G,'n'))
Out[7]: 
[('n', 'n2'),
 ('n', 'n3'),
 ('n', 'n1'),
 ('n2', 'n21'),
 ('n2', 'n22'),
 ('n1', 'n11'),
 ('n1', 'n13'),
 ('n1', 'n12'),
 ('n22', 'n221'),
 ('n13', 'n131')]

In [8]: [t for (s,t) in networkx.bfs_edges(G,'n')]
Out[8]: ['n2', 'n3', 'n1', 'n21', 'n22', 'n11', 'n13', 'n12', 'n221', 'n131']

In [9]: networkx.single_source_shortest_path_length(G,'n')
Out[9]: 
{'n': 0,
 'n1': 1,
 'n11': 2,
 'n12': 2,
 'n13': 2,
 'n131': 3,
 'n2': 1,
 'n21': 2,
 'n22': 2,
 'n221': 3,
 'n3': 1}

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