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Finding supernodes with insufficient frame size

Adam Holley <holl...@...>
 

Using cassandra as the backend, I was trying to count edges {g.E().count} and of course ran into the framesize problem because I had a supernode.  I found that I could identify which node was the supernode with:

g.V().toList().each{it.println(it);g.V(it).outE().count()}

This will print out the supernode vertex right before the framesize exception.  Then I can gradually increase the framesize until g.V(supernodeID) doesn't cause an exception.

Is there a better way to find supernodes without just increasing framesize to a large number, finding the supernode, and then working backward by decreasing the framesize?
As I'm not as familiar with Cassandra, once I know the supernode ID, is there a way to use CQL to determine the appropriate framesize without just gradually increasing or decreasing in the config?
Robert Dale <rob...@...>
 


This should give you the counts, highest first, by vertex id:

g.V().group().by(id()).by(outE().count()).order(local).by(values,decr)



Robert Dale

On Wed, Jun 14, 2017 at 11:32 AM, Adam Holley <holl...@...> wrote:
Using cassandra as the backend, I was trying to count edges {g.E().count} and of course ran into the framesize problem because I had a supernode.  I found that I could identify which node was the supernode with:

g.V().toList().each{it.println(it);g.V(it).outE().count()}

This will print out the supernode vertex right before the framesize exception.  Then I can gradually increase the framesize until g.V(supernodeID) doesn't cause an exception.

Is there a better way to find supernodes without just increasing framesize to a large number, finding the supernode, and then working backward by decreasing the framesize?
As I'm not as familiar with Cassandra, once I know the supernode ID, is there a way to use CQL to determine the appropriate framesize without just gradually increasing or decreasing in the config?

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Adam Holley <holl...@...>
 

That won't work if the framesize is not large enough.


On Wednesday, June 14, 2017 at 10:51:09 AM UTC-5, Robert Dale wrote:

This should give you the counts, highest first, by vertex id:

g.V().group().by(id()).by(outE().count()).order(local).by(values,decr)



Robert Dale
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