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Re: [igraph] On walktrap clustering and isolated vertices


From: Tamás Nepusz
Subject: Re: [igraph] On walktrap clustering and isolated vertices
Date: Tue, 27 May 2014 22:54:03 +0200

> I'm a bit confused. I tried this on an empty graph with ten vertices, and
> it puts all of the vertices in the same cluster.
No, it completes the dendrogram first, and then when you convert it to a flat 
clustering using the .as_clustering() method, it tries to decide where to "cut" 
the dendrogram by looking at the modularity scores of the clusterings after 1, 
2, ..., n merges. Since the modularity score is zero in your case for all 
possible cuts, it will arbitrarily choose one (depending on how I wrote the 
.as_clustering() method ;)), and it looks like it decides to stop after all the 
merges have been performed.

> Similarly, if I have a K_9 with a single isolated vertex, I also get one big 
> cluster.
Interestingly enough, the algorithm is not incorrect here either. If you put 
every single node into one big cluster, you get a modularity of zero. If you 
put the 9 nodes in the K_9 part into one cluster and the isolated vertex into 
another, you still get a modularity of zero. However, any other configuration 
during the merges performed by the walktrap clustering algorithm yields a 
negative modularity score.

> Printing the dendrogram after running it through the
> fixing function still causes an exception, though now it looks like a
> Python 3 compatibility issue
Yes, it is -- I'll try to fix that ASAP.

Actually, there's still one issue with the fix_dendrogram function but you may 
have figured that out already; the following statement is missing from the end:

cl._nitems = graph.vcount()

-- 
T.



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