.select("") and .as("") behavior with .where("")

[Alberto Pelosi]

Hi,

I have a question about .select("") and .as("") with .where("")

Suppose to have:

 g.V().hasLabel("x").as("xAlias").
    out("y").where(someCondition).select("xAlias")
   
   
Using .select("xAlias") I expect to obtain what I have labeled with "xAlias", that is all the vertex with label "x".

On the contrary I obtain all the vertex with label "x" that satisfy "out("y").where(someCondition)"

Is this a bug or I misunderstand .select()  and .as() behavior ?

Alberto

[Kelvin Lawrence]

Hi Alberto,

When you issue a query such as yours, you cause a number of traversers to be spawned. For each one that meets the criteria of the 'where' step you will get an instance of 'xAlias' as the result. So if the 'where' matches 10 times you will get 10 copies in the result.  

If you add a ".dedup()" to the end of your query you will just get one but this is not particularly useful.

The key thing is to think about how the query spawns traversers behind the scenes.

Hope that helps

Kelvin

[Kelvin Lawrence]

One more suggestion - try adding a ".path()" to the end of your query and you will hopefully see all the paths that the traversers took.

Cheers

Kelvin

[Daniel Kuppitz]

Instead of traversing all the paths and then deduping all the xAliases, you can also do:

g.V().hasLabel("x").filter(out("y").where(...))

This is basically just a hasNext() on the out("y").where(...) part.

Cheers,

Daniel

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[Robert Dale]

'as()' just labels parts in a path.

It's probably best described here - http://tinkerpop.apache.org/docs/current/reference/#path-data-structure

The result is the fully traversed path that meets the conditions of the traversal.

Even if you do g.V().hasLabel("x"), you're not really getting all 'x' in one set. Instead you're getting each individual 'x' as its own traverser.

If you really wanted ALL 'x' in one group, you would use something like aggregate().

g.V().hasLabel("x").aggregate("xAlias").out("y").where(someCondition).select("xAlias")

Now, 'xAlias' would contain the entire set of 'hasLabel('x'). Then of course the 'out().where()' was pointless.

Robert Dale

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[Alberto Pelosi]

Thank you for your answers, now it's clear.

My comprehension of "select" and "as" was not so deep.

My question arise from this use case:

- suppose you have some vertices with label "x" and a property "k"

- suppose some of them have __.out("edgeLabel1") (but not __.out("edgeLabel2"), by design)
    => __.where(__.out("edgeLabel1")). We can call them "firstSet"

- suppose some of them have __.out("edgeLabel2") (but not __.out("edgeLabel1"), by design)
    => __.where(__.out("edgeLabel2")). We can call them "secondSet"
   
   
How can I extract in one traversal the intersection between the two sets based on the value
of the "k" property?

My attemp was something like:

g.V().hasLabel("x").as("allX").
where(__.out("edgeLabel1")).as("firstSet").values("k").as("firstSetKValue").
select("allX").where(__.out("edgeLabel2")).as("secondSet").values("k").as("secondSetKValue").
select("firstSetKValue","secondSetKValue").where("firstSetKValue", P.eq("secondSetKValue"))

Now it's clear why it doesn't work, but there is a way to extract this information in one traversal?

Alberto

Robert Dale

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[Daniel Kuppitz]

Not sure whether you're looking for the k values or the x vertices in your result, so here are 2 queries to cover both cases:

x vertices:

g.V().hasLabel("x").

  sideEffect(out("edgeLabel1").aggregate("k").by("k")).

  barrier().

  filter(out("edgeLabel2").values("k").where(within("k")))

k values:

g.V().hasLabel("x").

  sideEffect(out("edgeLabel1").aggregate("k").by("k")).

  barrier().

  out("edgeLabel2").values("k").where(within("k")).

  dedup()

Cheers,

Daniel

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[Alberto Pelosi]

Thank you Daniel,

the second solution works well!
The first one, I don't know why, seems to ignore the condition "__.where(within("k")".

Thank you a lot,

Alberto