On Mon, Jan 6, 2014 at 5:48 PM, Merton Lister <[email protected]> wrote:
> Hi Csardi, > > Thanks for your suggestion. I looked at the formula on page 19 of the > manual, and it seems that the based on that formula the assortativity > should be 1 if all the edges in a (undirected) graph connect vertices of > the same type. > > So why it is returning 'NaN' in my case? Or is my understanding wrong? > Thanks. > If there is only one vertex type, then there is zero in the denominator, for both formulas. That's why you get NaNs. If there is only one vertex type, the question "do vertices connect preferentially to the same type?" does not make sense. Gabor > > Best regards, > > Merton > > >
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