Gremlin-python subGraph becomes a dict instead of a graph.

[Olav Laudy]

Hi,

Gremlin 3.3.1 with corresponding Gremlin-Python, playing with the Modern-graph.

g.E().hasLabel('knows').subgraph('subGraph').cap('subGraph').next() gives:

{u'@type': u'tinker:graph',
 u'@value': {u'edges': [e[7][1-knows->2], e[8][1-knows->4]],
  u'vertices': [v[1], v[2], v[4]]}}

Per console instruction:

gremlin> subGraph = g.E().hasLabel('knows').subgraph('subGraph').cap('subGraph').next() 
\
==>tinkergraph[vertices:3 edges:2]
gremlin> sg = subGraph.traversal()
==>graphtraversalsource[tinkergraph[vertices:3 edges:2], standard]

I cast to Python:

subGraph= g.E().hasLabel('knows').subgraph('subGraph').cap('subGraph').next()
sg = subGraph.traversal()

to receive:

AttributeError: 'dict' object has no attribute 'traversal'

How do I create a graph object from this so I can do subsequent queries, as per console:

gremlin> sg.E()

Thanks!

[Stephen Mallette]

subgraph() isn't well supported in the GLVs. i've made a note to improve documentation there. the problem is that GLVs deal with "reference" graph elements (elements that don't have properties) and there is no Graph class in GLVs to convert the returned subgraph to. I think you would want to return edge lists instead - I think that's the best workaround that i can think of.

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