Question 18693 in Amazon Web Sevices Question posted in
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Gremlin Python(AWS Neptune) get vertices with relations to list of other vertices – Amazon Web Sevices

New to Gremlin_python and GraphDBs in general, we are using AWS neptune. I am trying to get a list of scenarios that have relations to a set of different tags e.g. give me all scenarios that has tag A and tag B. But the set of tags can be dynamic(from 1 to 10 elements).

The question as i understand it in GraphDB therms is:

Give me all vertices that has an outgoing edge with the label=tag to vertices with label A AND B. Picture of the Graph

For a none dynamic approach i get it to work with this guery:

e = g.V().and_(__.out('tag').hasLabel('A'), __.out('tag').hasLabel('B')).label().toList()

But i am not able to get this to work with a list of tags. I also did try this but it just generate an OR statement and i have not found a way to get it to do and:

label_list = ['A','B']
e = g.V().where(__.out('has tag').hasLabel((*label_list))).label().toList()

I also tried this approach with just getting the message "Connection was closed by server", i have seen this many times with AWS Neptune that instead of generating an error message a get connection closed instead anyone know way this happens?:

label_list = ['A','B']
e = g.V().and_(map(lambda x:__.out('has tag').hasLabel(x), label_list)).label().toList()
print(len(e))

We are testing this in the sage maker notebook using python 3.8.

So internet can you please generate an answer for me here?

BR
Jon

List of includes in the notebook:

import yaml
from pathlib import Path
import json
import os

#from __future__  import print_function  # Python 2/3 compatibility

from gremlin_python import statics
from gremlin_python.structure.graph import Graph
from gremlin_python.process.graph_traversal import __
from gremlin_python.process.strategies import *
from gremlin_python.driver.driver_remote_connection import DriverRemoteConnection
from gremlin_python.process.traversal import T
from gremlin_python.process.traversal import Order
from gremlin_python.process.traversal import Cardinality
from gremlin_python.process.traversal import Column
from gremlin_python.process.traversal import Direction
from gremlin_python.process.traversal import Operator
from gremlin_python.process.traversal import P
from gremlin_python.process.traversal import TextP
from gremlin_python.process.traversal import Pop
from gremlin_python.process.traversal import Scope
from gremlin_python.process.traversal import Barrier
from gremlin_python.process.traversal import Bindings
from gremlin_python.process.traversal import WithOptions

from argparse import ArgumentParser
from loguru import logger
import nest_asyncio
nest_asyncio.apply()
Tags:

2

Answers


  1. Chosen as BEST ANSWER
    0

    I did find one solution for the problem. More in a Python kind of way, than Gremlin. To flatten a list of code worked, but maybe there is a more readable solution.

    label_list = ['A','B']
e = g.V().and_(*[__.out('tag').hasLabel(x) for x in label_list]).label().toList()

    A net trick i did figure out here is that you can send a list of tuples to the code. So you can dynamically add ANDs for both the out() and hasLabel().

    label_list = [('tag','A'),('tag','B'),('app','C')]
e = g.V().and_(*[__.out(x[0]).hasLabel(x[1]) for x in label_list]).label().toList()

  2. 0

    You should be able to use the within predicate for this if the test comes down to "has any of these property values (tags)".

    g.V().out('tag').hasLabel(within('A','B','C'))

    UPDATED: Based on comments.

    If you need to ensure that from the starting vertex there is at least one edge to each of the desired label types, you can modify the query above as follows:

    g.V().where(out('tag').hasLabel(within('A','B','C')).
      label().dedup().count().is(3))

    UPDATED 2023-02-22

    In this particular case the within is not really needed, although the query optimizer should generate the same query plan regardless, The query could be written as:

    g.V().where(out('tag').hasLabel('A','B','C').
      label().dedup().count().is(3))
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