Apache TinkerPop•17mo ago

indexOf Vertex with given property in a sorted list

Hi all! I've been trying to create a paginated GET REST api and sometimes I need to enforce a range on the resulting vertices starting from a vertex matching a given ID. Let's say my gremlin query so far returns the following list of vertices, sorted based on some property comparison:

[
  V{id:Z}
  V{id:F}
  V{id:K}
  V{id:A}
  V{id:N}
  V{id:T}
  V{id:C}
]

[
  V{id:Z}
  V{id:F}
  V{id:K}
  V{id:A}
  V{id:N}
  V{id:T}
  V{id:C}
]

Is there a way of asking gremlin what is the index of vertex with id K in the above list? Can all of this be done in a single query so that my final result simply returns the following, given K and a range of 3?

[
  V{id:K}
  V{id:A}
  V{id:N}
]

[
  V{id:K}
  V{id:A}
  V{id:N}
]

Some pseudocode for clarity:

g.V()
.hasLabel("profile")
.order()
.by("date", gremlin.process.order.asc);   // at this point we have the whole list
.range(indexOf(id=K), indexOf(id=K) + 3)  // this is the hiccup,
                                          // how to find indexOf(id=K) in the sorted result
.toList()

g.V()
.hasLabel("profile")
.order()
.by("date", gremlin.process.order.asc);   // at this point we have the whole list
.range(indexOf(id=K), indexOf(id=K) + 3)  // this is the hiccup,
                                          // how to find indexOf(id=K) in the sorted result
.toList()

Thank you in advance

6 Replies

VividElites•17mo ago

Couldn't you just use the range step? https://tinkerpop.apache.org/docs/current/reference/#range-step I think you would have to know the range before hand. Accoridng to the docs, if you use -1 as the 2nd argument in .range() it will emit the reamining verticies So maybe something like .range(3, -1)

neuxregime•17mo ago

That's what I have in mind too, but I need to know the index of V{id:K} in advance so it would be something like

g.V()
.hasLabel("profile")
.order()
.by("date", gremlin.process.order.asc);   // at this point we have the whole list
.range(indexOf(id=K), indexOf(id=K) + 3)  // this is the hiccup,
                                          // how to find indexOf(id=K) in the sorted result
.toList()

g.V()
.hasLabel("profile")
.order()
.by("date", gremlin.process.order.asc);   // at this point we have the whole list
.range(indexOf(id=K), indexOf(id=K) + 3)  // this is the hiccup,
                                          // how to find indexOf(id=K) in the sorted result
.toList()

VividElites•17mo ago

Hmm im not sure

neuxregime•17mo ago

no worries, appreciate the help 🙏 @spmallette @KelvinL is this even possible with gremlin? I know this can be done with 2 queries: get the list and find the index of vertex with id, then submit another query with the correct range applied. But this is not really ACID compliant, that's why I'm looking for a single-query solution

kelvinl2816•17mo ago

If you have the results in a list, the Gremlin index step can give you the list back with each cell having an index as well as the value. So for example:

gremlin> g.inject(['A','B','C']).index()
==>[[A,0],[B,1],[C,2]]

gremlin> g.inject(['A','B','C']).index()
==>[[A,0],[B,1],[C,2]]

Using that building block you should be able to build what you need. Often the limit and tail steps are useful when working with indexed lists. So if we have

gremlin> g.V(1,2,3).fold().index().unfold()
==>[v[1],0]
==>[v[2],1]
==>[v[3],2]

gremlin> g.V(1,2,3).fold().index().unfold()
==>[v[1],0]
==>[v[2],1]
==>[v[3],2]

we might do

gremlin> g.V(1,2,3).fold().index().unfold().where(limit(local,1).id().is(1))
==>[v[1],0]

gremlin> g.V(1,2,3).fold().index().unfold().where(limit(local,1).id().is(1))
==>[v[1],0]

and then

gremlin> g.V(1,2,3).fold().index().unfold().where(limit(local,1).id().is(1)).tail(local,1)
==>0

gremlin> g.V(1,2,3).fold().index().unfold().where(limit(local,1).id().is(1)).tail(local,1)
==>0

Similarly

gremlin> g.V(1,2,3).fold().index().unfold().where(limit(local,1).id().is(2)).tail(local,1)
==>1

gremlin> g.V(1,2,3).fold().index().unfold().where(limit(local,1).id().is(2)).tail(local,1)
==>1

neuxregime•17mo ago

As usual, this is brilliant Kelvin. I'll put this into work and report back 🙏

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indexOf Vertex with given property in a sorted list