Gremlin graph traversal that uses previous edge property value to filter later edges
In a graph traversal I only want to consider edges that have a property that is equal to the property of one of the edges visited in a previous step in the traversal.
I found http://tinkerpop.apache.org/docs/current/recipes/#traversal-induced-values but this appears to only work for a single object, in my case I need the value to change as I traverse. For example starting at V1 that has the outbound edges (E1, E2, E3...) I want to traverse out E1 to V2 and then traverse along any edge from V2 where edge.property(x) == E1.property(x), and do the same for all edges out of V1 (E2, E3, ...)
I can't find any documentation that supports a way to do this in Gremlin, is it possible?
gremlin graph-traversal
asked Oct 20 '16 at 22:18
Chris ReevesChris Reeves
313
add a comment |
In a graph traversal I only want to consider edges that have a property that is equal to the property of one of the edges visited in a previous step in the traversal.
I found http://tinkerpop.apache.org/docs/current/recipes/#traversal-induced-values but this appears to only work for a single object, in my case I need the value to change as I traverse. For example starting at V1 that has the outbound edges (E1, E2, E3...) I want to traverse out E1 to V2 and then traverse along any edge from V2 where edge.property(x) == E1.property(x), and do the same for all edges out of V1 (E2, E3, ...)
I can't find any documentation that supports a way to do this in Gremlin, is it possible?
gremlin graph-traversal
asked Oct 20 '16 at 22:18
Chris ReevesChris Reeves
313
add a comment |
In a graph traversal I only want to consider edges that have a property that is equal to the property of one of the edges visited in a previous step in the traversal.
I found http://tinkerpop.apache.org/docs/current/recipes/#traversal-induced-values but this appears to only work for a single object, in my case I need the value to change as I traverse. For example starting at V1 that has the outbound edges (E1, E2, E3...) I want to traverse out E1 to V2 and then traverse along any edge from V2 where edge.property(x) == E1.property(x), and do the same for all edges out of V1 (E2, E3, ...)
I can't find any documentation that supports a way to do this in Gremlin, is it possible?
gremlin graph-traversal
asked Oct 20 '16 at 22:18
Chris ReevesChris Reeves
313
In a graph traversal I only want to consider edges that have a property that is equal to the property of one of the edges visited in a previous step in the traversal.
I found http://tinkerpop.apache.org/docs/current/recipes/#traversal-induced-values but this appears to only work for a single object, in my case I need the value to change as I traverse. For example starting at V1 that has the outbound edges (E1, E2, E3...) I want to traverse out E1 to V2 and then traverse along any edge from V2 where edge.property(x) == E1.property(x), and do the same for all edges out of V1 (E2, E3, ...)
I can't find any documentation that supports a way to do this in Gremlin, is it possible?
gremlin graph-traversal
gremlin graph-traversal
asked Oct 20 '16 at 22:18
Chris ReevesChris Reeves
313
asked Oct 20 '16 at 22:18
Chris ReevesChris Reeves
313
edited Oct 20 '16 at 22:37
Chris Reeves
asked Oct 20 '16 at 22:18
Chris ReevesChris Reeves
313
asked Oct 20 '16 at 22:18
Chris ReevesChris Reeves
313
asked Oct 20 '16 at 22:18
Chris ReevesChris Reeves
313
313
add a comment |
add a comment |
1 Answer
1
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We can use the modern toy graph that ships with TinkerPop:
gremlin> graph = TinkerFactory.createModern()
==>tinkergraph[vertices:6 edges:6]
gremlin> g = graph.traversal()
==>graphtraversalsource[tinkergraph[vertices:6 edges:6], standard]
This graph already contains 6 edges of which two have a weight of 1.0:
gremlin> g.E().valueMap()
==>[weight:0.5]
==>[weight:1.0]
==>[weight:0.4]
==>[weight:1.0]
==>[weight:0.4]
==>[weight:0.2]
We use one of these two edges with a weight of 1.0 to get the other edge with the same weight. By traversing the first of these two edges, we land at a vertex that has two outgoing edges:
gremlin> g.V(1).outE().has('weight',1.0).inV().outE()
==>e[10][4-created->5]
==>e[11][4-created->3]
gremlin> g.V(1).outE().has('weight',1.0).inV().outE().valueMap()
==>[weight:1.0]
==>[weight:0.4]
One of these edges is the other one with the weight of 1.0. So, we only have to filter these edges based on the weight of the first edge:
gremlin> g.V(1).outE().has('weight',1.0).
as('firstEdge'). // save the first edge
inV().outE().
where(eq('firstEdge')). // compare with the first edge
by('weight') // use only the 'weight' property for the equality check
==>e[10][4-created->5]
answered Oct 21 '16 at 7:24
Florian HockmannFlorian Hockmann
1,439717
1
Is there any way to accomplish this without the match() step? Microsoft Azure's CosmosDB supports the gremlin query language, expect it can't handle the match() step yet :(
– SnoopDougg
Nov 14 '18 at 14:08
Thanks for asking about thematch()step. While reading my answer again, I noticed that it isn't really necessary to usematch()here. I edited my answer to solve the problem without usingmatch()which makes it much more readable in my opinion.
– Florian Hockmann
Nov 21 '18 at 17:00
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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active
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votes
We can use the modern toy graph that ships with TinkerPop:
gremlin> graph = TinkerFactory.createModern()
==>tinkergraph[vertices:6 edges:6]
gremlin> g = graph.traversal()
==>graphtraversalsource[tinkergraph[vertices:6 edges:6], standard]
This graph already contains 6 edges of which two have a weight of 1.0:
gremlin> g.E().valueMap()
==>[weight:0.5]
==>[weight:1.0]
==>[weight:0.4]
==>[weight:1.0]
==>[weight:0.4]
==>[weight:0.2]
We use one of these two edges with a weight of 1.0 to get the other edge with the same weight. By traversing the first of these two edges, we land at a vertex that has two outgoing edges:
gremlin> g.V(1).outE().has('weight',1.0).inV().outE()
==>e[10][4-created->5]
==>e[11][4-created->3]
gremlin> g.V(1).outE().has('weight',1.0).inV().outE().valueMap()
==>[weight:1.0]
==>[weight:0.4]
One of these edges is the other one with the weight of 1.0. So, we only have to filter these edges based on the weight of the first edge:
gremlin> g.V(1).outE().has('weight',1.0).
as('firstEdge'). // save the first edge
inV().outE().
where(eq('firstEdge')). // compare with the first edge
by('weight') // use only the 'weight' property for the equality check
==>e[10][4-created->5]
answered Oct 21 '16 at 7:24
Florian HockmannFlorian Hockmann
1,439717
1
Is there any way to accomplish this without the match() step? Microsoft Azure's CosmosDB supports the gremlin query language, expect it can't handle the match() step yet :(
– SnoopDougg
Nov 14 '18 at 14:08
Thanks for asking about thematch()step. While reading my answer again, I noticed that it isn't really necessary to usematch()here. I edited my answer to solve the problem without usingmatch()which makes it much more readable in my opinion.
– Florian Hockmann
Nov 21 '18 at 17:00
add a comment |
We can use the modern toy graph that ships with TinkerPop:
gremlin> graph = TinkerFactory.createModern()
==>tinkergraph[vertices:6 edges:6]
gremlin> g = graph.traversal()
==>graphtraversalsource[tinkergraph[vertices:6 edges:6], standard]
This graph already contains 6 edges of which two have a weight of 1.0:
gremlin> g.E().valueMap()
==>[weight:0.5]
==>[weight:1.0]
==>[weight:0.4]
==>[weight:1.0]
==>[weight:0.4]
==>[weight:0.2]
We use one of these two edges with a weight of 1.0 to get the other edge with the same weight. By traversing the first of these two edges, we land at a vertex that has two outgoing edges:
gremlin> g.V(1).outE().has('weight',1.0).inV().outE()
==>e[10][4-created->5]
==>e[11][4-created->3]
gremlin> g.V(1).outE().has('weight',1.0).inV().outE().valueMap()
==>[weight:1.0]
==>[weight:0.4]
One of these edges is the other one with the weight of 1.0. So, we only have to filter these edges based on the weight of the first edge:
gremlin> g.V(1).outE().has('weight',1.0).
as('firstEdge'). // save the first edge
inV().outE().
where(eq('firstEdge')). // compare with the first edge
by('weight') // use only the 'weight' property for the equality check
==>e[10][4-created->5]
answered Oct 21 '16 at 7:24
Florian HockmannFlorian Hockmann
1,439717
1
Is there any way to accomplish this without the match() step? Microsoft Azure's CosmosDB supports the gremlin query language, expect it can't handle the match() step yet :(
– SnoopDougg
Nov 14 '18 at 14:08
Thanks for asking about thematch()step. While reading my answer again, I noticed that it isn't really necessary to usematch()here. I edited my answer to solve the problem without usingmatch()which makes it much more readable in my opinion.
– Florian Hockmann
Nov 21 '18 at 17:00
add a comment |
We can use the modern toy graph that ships with TinkerPop:
gremlin> graph = TinkerFactory.createModern()
==>tinkergraph[vertices:6 edges:6]
gremlin> g = graph.traversal()
==>graphtraversalsource[tinkergraph[vertices:6 edges:6], standard]
This graph already contains 6 edges of which two have a weight of 1.0:
gremlin> g.E().valueMap()
==>[weight:0.5]
==>[weight:1.0]
==>[weight:0.4]
==>[weight:1.0]
==>[weight:0.4]
==>[weight:0.2]
We use one of these two edges with a weight of 1.0 to get the other edge with the same weight. By traversing the first of these two edges, we land at a vertex that has two outgoing edges:
gremlin> g.V(1).outE().has('weight',1.0).inV().outE()
==>e[10][4-created->5]
==>e[11][4-created->3]
gremlin> g.V(1).outE().has('weight',1.0).inV().outE().valueMap()
==>[weight:1.0]
==>[weight:0.4]
One of these edges is the other one with the weight of 1.0. So, we only have to filter these edges based on the weight of the first edge:
gremlin> g.V(1).outE().has('weight',1.0).
as('firstEdge'). // save the first edge
inV().outE().
where(eq('firstEdge')). // compare with the first edge
by('weight') // use only the 'weight' property for the equality check
==>e[10][4-created->5]
answered Oct 21 '16 at 7:24
Florian HockmannFlorian Hockmann
1,439717
We can use the modern toy graph that ships with TinkerPop:
gremlin> graph = TinkerFactory.createModern()
==>tinkergraph[vertices:6 edges:6]
gremlin> g = graph.traversal()
==>graphtraversalsource[tinkergraph[vertices:6 edges:6], standard]
This graph already contains 6 edges of which two have a weight of 1.0:
gremlin> g.E().valueMap()
==>[weight:0.5]
==>[weight:1.0]
==>[weight:0.4]
==>[weight:1.0]
==>[weight:0.4]
==>[weight:0.2]
We use one of these two edges with a weight of 1.0 to get the other edge with the same weight. By traversing the first of these two edges, we land at a vertex that has two outgoing edges:
gremlin> g.V(1).outE().has('weight',1.0).inV().outE()
==>e[10][4-created->5]
==>e[11][4-created->3]
gremlin> g.V(1).outE().has('weight',1.0).inV().outE().valueMap()
==>[weight:1.0]
==>[weight:0.4]
One of these edges is the other one with the weight of 1.0. So, we only have to filter these edges based on the weight of the first edge:
gremlin> g.V(1).outE().has('weight',1.0).
as('firstEdge'). // save the first edge
inV().outE().
where(eq('firstEdge')). // compare with the first edge
by('weight') // use only the 'weight' property for the equality check
==>e[10][4-created->5]
answered Oct 21 '16 at 7:24
Florian HockmannFlorian Hockmann
1,439717
edited Nov 21 '18 at 16:59
answered Oct 21 '16 at 7:24
Florian HockmannFlorian Hockmann
1,439717
answered Oct 21 '16 at 7:24
Florian HockmannFlorian Hockmann
1,439717
answered Oct 21 '16 at 7:24
Florian HockmannFlorian Hockmann
1,439717
1,439717
1
Is there any way to accomplish this without the match() step? Microsoft Azure's CosmosDB supports the gremlin query language, expect it can't handle the match() step yet :(
– SnoopDougg
Nov 14 '18 at 14:08
Thanks for asking about thematch()step. While reading my answer again, I noticed that it isn't really necessary to usematch()here. I edited my answer to solve the problem without usingmatch()which makes it much more readable in my opinion.
– Florian Hockmann
Nov 21 '18 at 17:00
add a comment |
1
Is there any way to accomplish this without the match() step? Microsoft Azure's CosmosDB supports the gremlin query language, expect it can't handle the match() step yet :(
– SnoopDougg
Nov 14 '18 at 14:08
Thanks for asking about thematch()step. While reading my answer again, I noticed that it isn't really necessary to usematch()here. I edited my answer to solve the problem without usingmatch()which makes it much more readable in my opinion.
– Florian Hockmann
Nov 21 '18 at 17:00
1
1
Is there any way to accomplish this without the match() step? Microsoft Azure's CosmosDB supports the gremlin query language, expect it can't handle the match() step yet :(
– SnoopDougg
Nov 14 '18 at 14:08
Is there any way to accomplish this without the match() step? Microsoft Azure's CosmosDB supports the gremlin query language, expect it can't handle the match() step yet :(
– SnoopDougg
Nov 14 '18 at 14:08
Thanks for asking about the
match() step. While reading my answer again, I noticed that it isn't really necessary to use match() here. I edited my answer to solve the problem without using match() which makes it much more readable in my opinion.– Florian Hockmann
Nov 21 '18 at 17:00
Thanks for asking about the
match() step. While reading my answer again, I noticed that it isn't really necessary to use match() here. I edited my answer to solve the problem without using match() which makes it much more readable in my opinion.– Florian Hockmann
Nov 21 '18 at 17:00
add a comment |
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