Using multiple variables for math.

[Sam Rosen]

Going off this example: http://tinkerpop.apache.org/docs/current/reference/#math-step

I'm trying to do some calculation for every vertex with the following code, but I continually get a null pointer exception. This seems to be my best attempt; where am I going wrong?

graph = TinkerFactory.createModern()
g = graph.traversalI()


g.V()
 .group()
 .by(
   both()
   .fold()
   .as("neighbors")
   .unfold()
    .outE()
    .fold()
    .as("edges")
    .math("neighbors - edges")
    .by(
         unfold().count())
    .by(
         unfold().count())
  )
  .by()
> java.lang.NullPointerException

Is there another way to accomplish arithmetic on some filtering of a traversal and the original traversals length?

Thanks,

Sam

[Robert Dale]

I believe the reason for the NPE is that there is a starting vertex where both().outE() does not exist and count() is applied to it.

Another thing is that you can reference back to variables prior to a reducing barrier. e.g. fold().as('a').unfold().fold().as('b').select('a') 

 

You have to account for cases where there are no results.

One way to do that is filter for where the full traversal exists:

g.V().where(__.both().outE()).groupCount().by(.....

Another option is to use count() which will always give a result:

gremlin> g.V().group().by(project('n','e').by(both().count()).by(both().outE().count()).math('n-e'))

==>[1.0:[v[1],v[6]],-2.0:[v[2]],0.0:[v[4]],-1.0:[v[5]],-3.0:[v[3]]]

There are probably other ways too.

Robert Dale

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[Sam Rosen]

My initial attempts all used counts but I'd still get null pointer exceptions or simply no right answers. The main problem with the given working option is that it requires computing the incident vertices twice.

[Robert Dale]

I see what you mean. Try this:

gremlin> g.V().group().by(both().fold().project('n','e').by(unfold().count()).by(unfold().outE().count()).math('n-e')).unfold()

==>[1.0:[v[1],v[6]],-2.0:[v[2]],0.0:[v[4]],-1.0:[v[5]],-3.0:[v[3]]]

Robert Dale

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[Sam Rosen]

Another good solution. Here's another one that allows maintaining the incident vertices with the result.

g.V()
.project("v", "neighbors")
.by()
.by(both().fold()) // find incident vertices
.group()
.by(
  project("count", "edge_count")
  .by(
    select("neighbors").unfold().count()
  )
  .by(
    select("neighbors").unfold().outE().count()
  ).math("count-edge_count")    
)
==>[1.0:[[v:v[1],n:[v[3],v[2],v[4]]],[v:v[6],n:[v[3]]]],-2.0:[[v:v[2],n:[v[1]]]],0.0:[[v:v[4],n:[v[5],v[3],v[1]]]],-1.0:[[v:v[5],n:[v[4]]]],-3.0:[[v:v[3],n:[v[1],v[4],v[6]]]]]