I have a graph where vertices have several outbound edge types and each of those edge types have properties. I want to get a count of outbound edges from each vertex but limit the count to edges with a certain label and property value.
So far I have:
g.V().hasLabel("vertexLabel").as("source")
.outE("edgeLabel").has("edgeProp", "propValue").as("edge")
.select("source", "edge")
This gives me a list of each source vertext and each outgoing edge with the correct label and property. What I want to do is reduce this to a single entry for each source vertex and count of the number of outbound edges. However I cannot seem to get groupCount to work in combination with outE. The closest I have got is:
g.V().hasLabel("vertexLabel").as("source").out("edgeLabel").groupCount()
This gives me count by source vertex but includes all the edges with that label regardless of what property values they have.
I know this probably needs a group().by() step but I am not sure how to form it.
Thanks in advance.
Using the "modern" toy graph from TinkerPop, I think you can best express this by using project():
gremlin> g.V().hasLabel('person').
......1> project('source','count').
......2> by().
......3> by(outE('created').has('weight',gt(0.5)).count())
==>[source:v[1],count:0]
==>[source:v[2],count:0]
==>[source:v[4],count:1]
==>[source:v[6],count:0]
You can do this with groupCount() as well but it feels a bit more awkward to me:
gremlin> g.V().hasLabel('person').
......1> outE('created').
......2> has('weight',gt(0.5)).
......3> groupCount().
......4> by(inV())
==>[v[5]:1]
Note you lose the "0" values in this case because the edges get filtered away prior to groupCount(). You could also go with group() but I don't think it reads as nicely as project():
gremlin> g.V().hasLabel('person').
......1> group().
......2> by().
......3> by(outE('created').
......4> has('weight',gt(0.5)).count())
==>[v[1]:0,v[2]:0,v[4]:1,v[6]:0]