2
$\begingroup$

I have a two-mode network which looks into relationship between forum threads. Two forum threads are connected if the same user has replied to both. The nodes are sized based on number of people who have only replied to the given forum thread, and no other ones (it would be the users who are unique to the given thread) and edges are the number of people who were active in both of the threads.

However, there are multiple problems with not doing anything with the edge weight:

1.) If there are two threads that are very active and have 500 users replying to them, and they have an edge of 100 between each other (100 common users), it shouldn't be as significant as if two less active forums share 100 common users;

2.) How do I solve the problem of a very active and a less active forum thread being connected, if at all, without using parallel edges?

What would be a good way to calculate significance of edge weight?

$\endgroup$
1
$\begingroup$

You might want to consider the approach outlined in Newman (2001) "Scientific collaboration networks. II. Shortest paths, weighted networks, and centrality." Physical Review E 64, 016132. It's implemented in the tnet package for R.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

The correct term for what you are calling "two-mode" network is bipartite network between users and threads. In particular, you are trying to analyze the projection on the threads of the bipartite network.

I suggest to apply a disparity filter, a network reduction algorithm to extract the backbone structure of (un)directed weighted networks. Disparity filter can sufficiently reduce the network without destroying the multi-scale nature of the network. Here you can find the paper that introduced the algorithm.

If you work with R, I have written a package that extracts the backbone structure of a graph with a given statistical significance. You can install it from CRAN.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.