I'm traversing a graph and need to obtain all leaf nodes sorted by an order property present on the edges.
I was only able to get the leaf vertices ordered by the last edge with the following query:
g.V('n1').repeat(out()).emit().order().by(inE().values('order'))
In this example I wish I could get something like [n3, n4, n5, n3] or a map with the accumulated order for every vertice ([n3: 1.1, n3: 3, n4: 1.2, n5: 2])
Your picture was nice but when asking questions about Gremlin it is best to add a small sample data script like this one:
g.addV().property(id,'n1').as('n1').
addV().property(id,'n2').as('n2').
addV().property(id,'n3').as('n3').
addV().property(id,'n4').as('n4').
addV().property(id,'n5').as('n5').
addV().property(id,'n6').as('n6').
addE('next').from('n1').to('n3').property('order',3).
addE('next').from('n1').to('n2').property('order',1).
addE('next').from('n1').to('n6').property('order',2).
addE('next').from('n2').to('n3').property('order',1).
addE('next').from('n2').to('n4').property('order',2).
addE('next').from('n2').to('n5').property('order',3)
I opted to provide an answer that gives a "map with the accumulated order for ever vertex" - well, not quite a "map" because the "n3" key would not be able to be used twice. It returns pairs where the first item is the leaf vertex and the second item is a list of the "order" values traversed along the way:
gremlin> g.V('n1').
......1> repeat(outE().inV()).
......2> emit(__.not(outE())).
......3> path().
......4> map(union(tail(local,1),
......5> unfold().values('order').fold()).
......6> fold())
==>[v[n3],[3]]
==>[v[n6],[2]]
==>[v[n3],[1,1]]
==>[v[n4],[1,2]]
==>[v[n5],[1,3]]
You can see I provided a condition to emit() to only produce the leaf vertices. I then transform the path() traversed using a map() which grabs the leaf vertex with tail(local,1)) for the first item in the pair and then unfolds the Path for "order" values for the second item in the pair.
