Python Networkx 中的社区检测
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/22070196/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Community detection in Networkx
提问by Alan Valejo
I'm studying about detection communities in networks.
我正在研究网络中的检测社区。
I'm use igraph and Python
我使用 igraph 和 Python
For the optimal number of communities in terms of the modularity measure:
对于模块化度量方面的最佳社区数量：
from igraph import *
karate = Nexus.get("karate")
cl = karate.community_fastgreedy()
cl.as_clustering().membership
For supply the desired number of communities:
为了提供所需数量的社区：
from igraph import *
karate = Nexus.get("karate")
cl = karate.community_fastgreedy()
k=2
cl.as_clustering(k).membership
However, I like to do this using networkx. I know get optimal number of communities in terms of the modularity measure:
但是，我喜欢使用 networkx 来做到这一点。我知道在模块化度量方面获得最佳社区数量：
import community # --> http://perso.crans.org/aynaud/communities/
import fastcommunity as fg # --> https://networkx.lanl.gov/trac/ticket/245
import networkx as nx
g = nx.karate_club_graph()
partition = community.best_partition(g)
print "Louvain Modularity: ", community.modularity(partition, g)
print "Louvain Partition: ", partition
cl = fg.communityStructureNewman(g)
print "Fastgreed Modularity: ", cl[0]
print "Fastgreed Partition: ", cl[1]
But I can not get the desired number of communities. Are there some algorithm for this, using Networkx?
但我无法获得所需数量的社区。是否有一些算法，使用 Networkx？
回答by zihaolucky
I'm also new to networkx and igraph, I used Gephi, an data visualization tool/software. And it has the same community detection algorithm as the one in networkx you are now using. Specifically, in http://perso.crans.org/aynaud/communities/
我也是 networkx 和 igraph 的新手，我使用了 Gephi，一种数据可视化工具/软件。它具有与您现在使用的 networkx 中相同的社区检测算法。具体来说，在http://perso.crans.org/aynaud/communities/
It uses the louvain method described in Fast unfolding of communities in large networks, Vincent D Blondel, Jean-Loup Guillaume, Renaud Lambiotte, Renaud Lefebvre, Journal of Statistical Mechanics: Theory and Experiment 2008(10), P10008 (12pp)
它使用在大型网络中社区的快速展开、Vincent D Blondel、Jean-Loup Guillaume、Renaud Lambiotte、Renaud Lefebvre、Journal of Statistical Mechanics: Theory and Experiment 2008(10)、P10008 (12pp) 中描述的 louvain 方法
You can not get desired number of communities, as I know, there're two ways worth to try:
您无法获得所需数量的社区，据我所知，有两种方法值得尝试：
- Use Gephi. You can use gephi and there's a parameter called
resolutionthat would change the size of the community you get. - Use NetworkX. This time, we may not use
best_partition(G)any more. But usepartition_at_level(dendrogram, level), I guess this might help.
- 使用 Gephi。您可以使用 gephi 并且有一个名为的参数
resolution可以改变您获得的社区的规模。 - 使用 NetworkX。这一次，我们可能不再使用
best_partition(G)了。但是使用partition_at_level(dendrogram, level)，我想这可能会有所帮助。
Check the source codehere for more info.
在此处查看源代码以获取更多信息。
回答by Johannes Wachs
Perhaps I am misunderstanding you, but if you would like the number of communities output by the NetworkX implementation of the best_partition algorithm, just note that best_partition(G) gives a dictionary with nodes as keys and their partition number as value.
也许我误解了您的意思，但是如果您想知道 best_partition 算法的 NetworkX 实现输出的社区数量，请注意 best_partition(G) 给出了一个以节点为键、分区号为值的字典。
You can count the number of unique values in a dictionary like this (likely not optimal):
您可以像这样计算字典中唯一值的数量（可能不是最佳的）：
dict = {'a':1,'b':1,'c':2,'d':1,'e':3,'f':4,'g':5}
count=list(set([i for i in dict.values()]))
print count
print len(count)
With result
结果
[1, 2, 3, 4, 5]
5
