Re: [igraph] Selecting edges by _source and _target in an undirected graph

[Tamas Nepusz]

Hi,

This is a bug in the Python interface; _source and _target do not work properly
on undirected graphs. As a workaround, use g.incident(x) to obtain the IDs of
the edges incident on vertex x -- this can then be used to subset g.es:

g.es[g.incident(x)]

Please file a bug report on https://github.com/igraph/python-igraph/issues/new
if you want to track the progress of this issue.

-- 
T.

On 05/06, Santiago Videla wrote:
> Hi,
> 
> 
> Using igraph-python, I found something which as far as I understand, it
> doesn't match the documentation:
> 
> 
> In [1]: import igraph as ig
> 
> In [2]: g = ig.Graph(3, [(0,1), (2,0)], directed=False)
> 
> In [3]: len(g.es.select(_source=0))
> 
> Out[3]: 2
> 
> In [4]: len(g.es.select(_target=0))
> 
> Out[4]: 0
> 
> Shouldn't g.es.select(_target=0) return 2 edges as well?
> 
> In fact, in the tutorial (
> http://igraph.org/python/doc/tutorial/tutorial.html#querying-vertices-and-edges-based-on-attributes)
> it says:
> 
> =============
> There are also a few special structural properties for selecting edges:
> 
> Using _source or _from in the keyword argument list of EdgeSeq.select()
> filters based on the source vertices of the edges. E.g., to select all the
> edges originating from Claire (who has vertex index 2):
> 
> >>> g.es.select(_source=2)
> 
> Using _target or _to filters based on the target vertices. This is
> different from _source and _from if the graph is directed.
> =============
> 
> I understand that for an undirected graph, it should hold::
> 
> g.es.select(_source=vid) == g.es.select(_target=vid)
> 
> or not?
> 
> Regards,
> 
> -- 
> Santiago Videla
> http://www.linkedin.com/in/svidela

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-- 
T.