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Re: [igraph] from Infomap tree output to communities object

From: Tamas Nepusz
Subject: Re: [igraph] from Infomap tree output to communities object
Date: Wed, 23 Mar 2016 18:21:14 +0100


I'm not that familar with R, but it seems that igraph has a
"make_clusters()" function in R that allows you to construct a
"communities" object from a graph and a membership vector where the
i-th element of the membership vector is simply the community index of
vertex i. So, if you can convert the output to a single R vector, the
rest should be easy.

Alternatively, make_clusters() can also take a "merge matrix" for
hierarchical community detection algorithms. A merge matrix is a
matrix with 2 columns and N rows and each row describes the merge of
two communities into a larger one. The initial vertices of the graph
are identified by integers 1..|V| in the matrix; then the community
that is created by the first merge is identified by |V|+1, the
community created by the second merge is identified by |V|+2 and so


On Wed, Mar 23, 2016 at 6:06 PM, Alexander Struck
<address@hidden> wrote:
> Thank you Tamas for looking into this.
> Has anybody else any pointer for me how to transform Infomap output into 
> something that I can use with igraph-R? I’d really like to keep using igraph 
> for further research in this project.
> Best regards,
> Alexander
>> On 23 Mar 2016, at 12:09, Tamas Nepusz <address@hidden> wrote:
>> Hi,
>> I don't think there is a built-in igraph function to convert the .tree
>> output format of the code on mapequation.org to an igraph communities
>> object.
>> T.
>> On Mon, Mar 21, 2016 at 6:51 PM, Alexander Struck
>> <address@hidden> wrote:
>>> Hi GuRus,
>>> I had to use the Infomap implementation from 
>>> http://mapequation.org/code.html and created a .tree text file. I must have 
>>> been using the wrong search terms but couldn’t find information on how to 
>>> turn that .tree structure into an igraph communities object. Which 
>>> transformation steps / options did I miss? I’d appreciate any hints.
>>> Many thanks,
>>> Alexander
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