recursive query with hierarchical result

[Greg H]

I have a graph structure representing a hierarchy of arbitrary depth, ie  

V -child-> V -child-> V -child-> V -child-> V ... 

I want to query a subset of the graph rooted at some V traversing all child edges recursively to some depth d, returning all found vertices and their properties in (ideally) a hierarchical structure. 

So for example given the graph

V(a) -child-> V(b)

V(a) -child-> V(c)

V(b) -child-> V(d)

V(c) -child-> V(e)

the query for the graph rooted at v(a) to depth 2 would look like

{ id:'a',

  keys:values,

  child: [{

    id:'b',

    keys:values,

    child:[{

     id:'d'

     keys:values

    },

    {

    id:'c',

    keys:values,

    child:[{

      id:'e',

      keys:values

    }]

  }]

}]

I'm new to gremlin and wondering if this is doable and how? 

[Daniel Kuppitz]

If you're using TP2, use the linked Wiki page. But if you're using TP3:

g.V("some V").repeat(out("child").simplePath().dedup()).times(depth).tree().next()

Cheers,

Daniel

On Thu, Jun 9, 2016 at 2:01 AM, Greg H <gr...@itactual.com> wrote:

Nevermind - https://github.com/tinkerpop/gremlin/wiki/Tree-Pattern seems to fit the bill nicely 

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[Greg H]

Thanks Daniel.

I'm on the TP2 stack and a simple tree() pattern query almost works, eg

g.V('x').out('child').out('child').tree().cap()  

This returns a subset of the graph rooted at x with a depth of 2, but it excludes paths with a depth less than 2 (ie child node without children) and I need to include those as well. To rephrase the requirement I want to select the subset of the tree rooted at x vertexes related by child to a depth up to and including d. 

Any suggestions on how to format that query in TP2? :)  

Cheers,

Greg

[Marko Rodriguez]

Hi Greg,

If you want both 0-step, 1-step, 2-step, etc. branches in your tree, simply use repeat() and emit().

gremlin> g = TinkerFactory.createModern().traversal()

==>graphtraversalsource[tinkergraph[vertices:6 edges:6], standard]

gremlin> g.V().repeat(out()).times(2).emit().tree()

==>[v[1]:[v[2]:[:], v[3]:[:], v[4]:[v[3]:[:], v[5]:[:]]], v[4]:[v[3]:[:], v[5]:[:]], v[6]:[v[3]:[:]]]

gremlin> g.V().repeat(out()).times(2).emit().tree().next()

==>v[1]={v[2]={}, v[3]={}, v[4]={v[3]={}, v[5]={}}}

==>v[4]={v[3]={}, v[5]={}}

==>v[6]={v[3]={}}

gremlin>

The times(2) is the upper limit and emit() says, “at every iteration of the walk, yield what has been touched thus far.

HTH,

Marko.

http://markorodriguez.com

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[Daniel Kuppitz]

This should be the equivalent TP2 query:

g.v("some V").as("x").out("child").simplePath().dedup().loop("x") {it.loops <= depth} {true}.tree().cap()

Cheers,

Daniel

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[Greg H]

Vero cool.  Still learning this stuff, thanks Marko

[Greg H]

Awesome thx. 

I'm in the process of upgrading 

[Greg H]

Mark,

Thanks for the help. 

The second form of the query is easier for me to process using a node client, so:

gremlin>g = TinkerFactory.createModern().traversal()

==>graphtraversalsource[tinkergraph[vertices:6 edges:6], standard]

gremlin> g.V(1).repeat(out()).times(2).emit().tree().next()

==>v[1]={v[2]={}, v[3]={}, v[4]={v[3]={}, v[5]={}}}

gremlin> g.V(4).repeat(out()).times(2).emit().tree().next()

==>v[4]={v[3]={}, v[5]={}}

gremlin> g.V(3).repeat(out()).times(2).emit().tree().next()

gremlin> 

The V(1) and V(4) results for this query are exactly what I want - the query root and the set of out traversals to depth of 2 inclusive - but the 0 step case returns an empty result. Would it be straightforward to return the query root and an empty traversal set, ie v[3]={} for this case? 

Thx again. 

  

On Thursday, June 9, 2016 at 7:23:05 AM UTC-6, Marko A. Rodriguez wrote:

[Daniel Kuppitz]

If the emit() comes before repeat(), elements will be emitted prior to the repeat() traversal being executed.

gremlin> g.V(1).emit().repeat(out()).times(2).tree().next()

==>v[1]={v[2]={}, v[3]={}, v[4]={v[3]={}, v[5]={}}}

gremlin> g.V(4).emit().repeat(out()).times(2).tree().next()

==>v[4]={v[3]={}, v[5]={}}

gremlin> g.V(3).emit().repeat(out()).times(2).tree().next()
==>v[3]={}

Cheers,

Daniel

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[Greg H]

Thats got it - tq Daniel