what is the weight of the network?
hello i have some problems about the function cna network
i use the trajectory dcd file to generate a amino acid network, and i saw that ,in the tutorials, Correlation network analysis can then be performed (with the *cna() function) to generate a correlation network with residues as nodes that are linked by weighted edges that are proportional to their degree of correlated motion. ** so i wander that how do the weights of the edges be caculated ?
when i increase the value of the cutoff.cij ,the network will filter out some edges with high value of weights, does that mean the weights of the edges are inversely proportional to the degree of correlated motion? or it is just the value of cij ?
in a general way , i think the larger value of weight is the better and the more improtant ,and at the time to caculate the shortest pathways ,does the caculation prefer the edges with larger value of weight? so ,if i want to caculate the shorest pathway , should i change the weight to reciprocal?
thank you very much
Comments (5)
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reporter its very helpful
thankyou verymuch
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reporter sorry , i had another question if the weight is proportional to -log(|cij|), how can i diffrentiate the completely correlated motions, C(ij) = 1, and the completely anticorrelated motions, C(ij) = - 1?
and , if there is a way to build a net contaning negative weight?
thankyou verymuch!
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Hi,
Unfortunately, we don't support negative correlation yet. Here is a related reference that might help: https://journals.aps.org/pre/abstract/10.1103/PhysRevE.80.016114.
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By default, the weight is inversely proportional to the correlation. Or, more precisely, proportional to
-log(|cij|)
. This is a conventional way to build up a "distance" network; See reference. By this definition, shortest path will pick up edges with lowest weights (because they represent "distance" between two nodes).You can of course build a network with edge weight representing "importance", by setting "minus.log=FALSE" in calling
cna()
.Let me know if you have more questions.