Well, if you want to get it duplicated, you can just do union(outE(), inE()),
that's easy and inexpensive. However, any way to get rid of the duplicates
can be expensive:
- local(bothE().dedup()) // needs to
keep track of all edges; requires internal memory structures
- union(outE(), __.as("a").inE().not(where(outV().as("a")))) // enables
partial path tracking; again that requires internal memory structures
If we / providers implement the deduplication though, we wouldn't require
any extra memory structures. Returning the duplicates makes sense to me
from an engineering perspective, but not from a user perspective.
Cheers,
Daniel
On Wed, Oct 25, 2017 at 6:53 PM, Robert Dale <robd...@gmail.com> wrote:
> I think that bothE() == union(outE(),inE()) and outE().count() +
> inE().count() == bothE().count(). If you don't want the self-referencing
> edge to be returned twice, then either make it a unidirected edge (if
> supported) so that it would still satisfy the two previous condition or
> dedup(). In either case, it's left to the user to determine what edges are
> returned.
>
> Also, I think it makes sense that g.E() == g.V().outE(). It should not be
> g.V().inE() due to potential for unidirected edges.
>
>
> Robert Dale
>
> On Wed, Oct 25, 2017 at 9:32 AM, Daniel Kuppitz <m...@gremlin.guru> wrote:
>
> > IMO it should return the edge only once.
> >
> > Cheers,
> > Daniel
> >
> >
> > On Wed, Oct 25, 2017 at 5:47 AM, Stephen Mallette <spmalle...@gmail.com>
> > wrote:
> >
> > > The test suite doesn't seem to enforce behavior related to
> self-relating
> > > edges. TinkerGraph does this:
> > >
> > > gremlin> g = TinkerGraph.open().traversal()
> > > ==>graphtraversalsource[tinkergraph[vertices:0 edges:0], standard]
> > > gremlin> g.addV().as('a').addE('self').to('a')
> > > ==>e[1][0-self->0]
> > > gremlin> g.E()
> > > ==>e[1][0-self->0]
> > > gremlin> g.V().bothE().count()
> > > ==>2
> > >
> > > Should bothE() return 2 in this case or 1? I think that we've said in
> the
> > > past that g.E() is the same as g.V().outE() or g.V().inE(), but not
> > > necessarily g.V().bothE(). Thoughts?
> > >
> >
>
