I want to quantify the effect of a node attribute on the tie formation. I've already done some analysis but not sure what is the best/right/proper way to do, so looking for some advice. Thanks !
This is a bipartite (two-mode) network :
Actors (MCs): 66 , Events (news): 7376, edges (non-weighted): 4938
There is an edge between an actor (Member of Congress) and an event (news) if MC tweets about (commentates on) that event.
This can be reduced to one-mode network:
A co-commentation network of congress members (MCs) (N=66, # of edges: 1863).
Each weighted edge represents the number of events incident MCs have commentated on Twitter.
Nodes have a single attribute: the political party they belong to.
What I've done so far:
1. Clustering (modularity based community detection) on the one-mode network. 95% of the MCs are found to be in the same group as their actual co-party members. So, this clearly indicates an effect of node attribute (party-match) on tie formation. But this does not look like the right way to quantify its effect from a statistical perspective?
2. Attempted ERGM on the bipartite network (using statnet) to see the effect of node match (when parties are not differentiated). Not sure what other parameters can be added. And how should I interpret ~0.32 here?
Summary of model fit
Formula: two_mode ~ edges + b1nodematch("party")
Iterations: 20 out of 20
Monte Carlo MLE Results:
Estimate Std. Error MCMC % p-value
edges -4.82148 0.02591 1 <1e-04 ***
b1nodematch.party 0.31572 0.02996 1 <1e-04 ***
Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1
Null Deviance: 674870 on 486816 degrees of freedom
Residual Deviance: 52736 on 486814 degrees of freedom
AIC: 52740 BIC: 52762 (Smaller is better.)
3. Now considering calculating the likelihood of tie formation with co-party members. For each MC mi I’ll get (sum of edge weights with co-party members) / (weighted node degree*) and then average them.
* : i.e., sum of all the edge weights mi incident to
Any pointers and feedbacks are more than welcome.
PS. Here is the IPYNB.