使用gremlin-python过滤列表的边缘作为属性值

Question

I'm storing a list as the property value of some edges in my graph, similar to the question asked here . The solution to that question was given in JavaScript, but I'm looking for a way to do the same thing in Python.

Also, note that Amazon Neptune doesn't support Lambda steps , so the solution can't use lambdas.

Answer 1

I'm not sure if this is still an issue to you or not, but I wonder if this wouldn't work for you:

gremlin> g = TinkerGraph.open().traversal()
==>graphtraversalsource[tinkergraph[vertices:0 edges:0], standard]
gremlin> g.addV('person').as('a').addE('self').to('a').property('x',[1,2,3]).select('a').addE('self').to('a').property('x',[10,20,30]).iterate()
gremlin> g.E().valueMap(true)
==>[x:[1,2,3],id:3,label:self]
==>[x:[10,20,30],id:4,label:self]
gremlin> g.V().outE('self').filter(local(values('x').unfold().is(2))).valueMap(true)
==>[x:[1,2,3],id:3,label:self]

Basically you could do a local() unfolding of your list property and then filter on that. Neptune likely won't optimize this filter (eg use an index) but if you aren't filtering a large number of edges it might be a sufficient workaround (if it works).

Note that I wrote the above in Groovy so as to test it easily. In Python, you must pay attention to certain minor syntax modifications around naming conflicts. Therefore, the last traversal there would be written like this in python (because is is a reserved word in python):

g.V().outE('self').filter(local(values('x').unfold().is_(2))).valueMap(true)