Thanks for the clarification and the quick modification. Snehal On Wed, Oct 4, 2017 at 7:04 PM, Tiago de Paula Peixoto <tiago@skewed.de> wrote:
On 04.10.2017 14:26, Snehal Shekatkar wrote:
Thanks for the quick reply. It is indeed true that variance should be NaN but assortativity would be zero if I understand it correctly.
No, that is not correct. The variance is not NaN, it is zero. Just look at the formula for scalar assortativity: If the variance is zero, the value of the coefficient is undefined, as I explained.
Now, when instead of 'float', I use 'int' as the type for the property map, I do get 0 value for the assortativity. Thus I guess that the values are wrong and it is a bug. Am I right? From your reply, it isn't clear to me if this is a bug.
It is not a bug. The reason why you get different answers is due to numerical instability. Instead of a variance of zero, what ends up computed instead is a very small number due to limited numerical precision.
I've modified the version in git to always return NaN in such cases, instead of this unstable behavior.
-- Tiago de Paula Peixoto <tiago@skewed.de>
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-- Snehal M. Shekatkar Pune India