fold initial value derived from the graph

[Bailey Glen]

How do you provide an initial value to fold that is derived from the traversal. For instance

g.V().values('pval').flatmap(

  _as('cur_val')

  .union(flatMap( <generate some value>

  )).fold(select('cur_val'), sum))

[Daniel Kuppitz]

Why not simplify your query and just use sum()? E.g. instead of

g.V(1).as("a").out("knows").values("age").fold(select("a").by("age"), sum)

...you could do:

gremlin> g.V(1).as("a").union(identity(), out("knows")).values("age").sum()

==>88

Cheers,

Daniel

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[Bailey Glen]

I used sum as a simplified example, I'm actually attempting to apply a lambda function to perform some calculations.

Is it possible to have the initial fold value derived from the graph?

On Friday, October 13, 2017 at 3:36:28 PM UTC-4, Daniel Kuppitz wrote:

Why not simplify your query and just use sum()? E.g. instead of

g.V(1).as("a").out("knows").values("age").fold(select("a").by("age"), sum)

...you could do:

gremlin> g.V(1).as("a").union(identity(), out("knows")).values("age").sum()

==>88

Cheers,

Daniel

On Fri, Oct 13, 2017 at 9:56 AM, Bailey Glen <baile...@gmail.com> wrote:

How do you provide an initial value to fold that is derived from the traversal. For instance

g.V().values('pval').flatmap(

  _as('cur_val')

  .union(flatMap( <generate some value>

  )).fold(select('cur_val'), sum))

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[Daniel Kuppitz]

Is it possible to have the initial fold value derived from the graph?

No, fold() can only have a constant / static initial value, but you can always use a map() lambda step and do whatever you want with all the information available on the current traverser.

Cheers,

Daniel

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