Get different label vertex for group aggregation based on date

Question

    v1=graph.addVertex(label,"l1","submit_time",Fri Apr 26 21:01:36 PDT 2019) //v[2345432]
    v2=graph.addVertex(label,"l2","start_time",Fri Apr 26 22:01:36 PDT 2019) // v[409632904]
    v3=graph.addVertex(label,"l2","start_time",Fri Apr 26 22:01:36 PDT 2019)  //v[204824704]
    v4=graph.addVertex(label,"l2","start_time",Fri Apr 26 23:01:36 PDT 2019). //v[307241008]

    Edge e1 = v1.addEdge("e1", v2);
    Edge e2 = v1.addEdge("e1", v3);
    Edge e3 = v1.addEdge("e1", v4);




    g.V().hasLabel("l2").group().by(map{(it.get().value("start_time").getYear()+1900)+"/"+(it.get().value("start_time").getMonth()+1)+"/"+it.get().value("start_time").getDate()+" "+it.get().value("start_time").getHours()})

we are getting below output: Output1: 2019/4/26 23:[v[307241008]], 2019/4/26 22:[v[409632904],v[204824704]]

Can anyone please help me to get for each of the aggregated values(aggregated by l2 and all the l2 vertex has edge to l1 ), so I need to get its corresponding l1 label vertex also in single query. eg : Output2: 2019/4/26 23:[v[307241008]], v[2345432] 2019/4/26 22:[v[409632904],v[204824704]] ,v[2345432] Thanks.

I don't see a question here. Can you maybe add a little more context around what you need help with? — Aaron, May 07 '19 at 13:17

score 3 · Accepted Answer · answered May 07 '19 at 16:42

Let me start with a proper script to create the sample graph, so others can easier follow along:

g = TinkerGraph.open().traversal()
g.addV('l1').
    property(id, 2345432).
    property('submit_time', new Date('Fri Apr 26 21:01:36 PDT 2019')).
  addV('l2').
    property(id, 409632904).
    property('start_time', new Date('Fri Apr 26 22:01:36 PDT 2019')).
  addV('l2').
    property(id, 204824704).
    property('start_time', new Date('Fri Apr 26 22:01:36 PDT 2019')).
  addV('l2').
    property(id, 307241008).
    property('start_time', new Date('Fri Apr 26 23:01:36 PDT 2019')).
  addE('e1').from(V(2345432)).to(V(409632904)).
  addE('e1').from(V(2345432)).to(V(204824704)).
  addE('e1').from(V(2345432)).to(V(307241008)).iterate()

And your query properly formatted:

g.V().hasLabel("l2").
  group().
    by {(it.value("start_time").getYear() + 1900) + "/" +
        (it.value("start_time").getMonth() + 1) + "/" +
         it.value("start_time").getDate() + " " +
         it.value("start_time").getHours()}

Now, if you want to add all the l1 vertices, you can no longer use a simple Map for you result. Each entry needs its own map, so you can capture a third field. Thus, you need to unfold the map and reshape it with a project() step:

g.V().hasLabel("l2").
  group().
    by {(it.value("start_time").getYear() + 1900) + "/" +
        (it.value("start_time").getMonth() + 1) + "/" +
         it.value("start_time").getDate() + " " +
         it.value("start_time").getHours()}.
  unfold().
  project('time','l2','l1').
    by(keys).
    by(values).
    by(select(values).unfold().in('e1').dedup().fold())

This will yield:

gremlin> g.V().hasLabel("l2").
......1>   group().
......2>     by {(it.value("start_time").getYear() + 1900) + "/" +
......3>         (it.value("start_time").getMonth() + 1) + "/" +
......4>          it.value("start_time").getDate() + " " +
......5>          it.value("start_time").getHours()}.
......6>   unfold().
......7>   project('time','l2','l1').
......8>     by(keys).
......9>     by(values).
.....10>     by(select(values).unfold().in('e1').dedup().fold())
==>[time:2019/4/26 23,l2:[v[307241008]],l1:[v[2345432]]]
==>[time:2019/4/26 22,l2:[v[409632904],v[204824704]],l1:[v[2345432]]]

Thank you! @Daniel Kuppitz..i got required output with your query — User12345, May 08 '19 at 04:43

Get different label vertex for group aggregation based on date

1 Answers1