How to query nodes with multiple relationships? - Neo4j

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Message from discussion How to query nodes with multiple relationships?
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From: Michael Hunger <michael.hun...@neotechnology.com>
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Subject: Re: [Neo4j] How to query nodes with multiple relationships?
Date: Fri, 10 Aug 2012 19:13:09 +0200
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you can write the first one like this:

> start a=3Dnode(1405) match a-[:A*1..3]->m
> where a-[:B*1..3]->m
> return distinct m.name

Please remember that distinct is an expensive operation (in term of =
non-lazyness and memory consumption) as it has to fetch all data and =
merge the results in memory.

You may at least try to run it without distinct to see the difference.

Michael

Am 10.08.2012 um 18:59 schrieb fli.rpx:

> Hi Michael,
>=20
> Thanks for your reply.
>=20
> There are differences between the two queries. The first query, "start =
a=3Dnode(1405) match a-[:A*1..3]->m, a-[:B*1..3]->m return distinct =
m.name", has AND relationship. Its result set will be the intersection =
of two traversals.
>=20
> The second query, "start a=3Dnode(1405) match a-[:A|B*1..3] return =
distinct m; ", is OR relationship.=20
>=20
> These two queries have performance difference. The first one will take =
two traversals then compare. The second one only has one traversal.
>=20
> I am wondering if I can write the first query in a way of the second =
query. Something like this?:
>=20
> start a=3Dnode(1405) match a-[:A&B*1..3] return distinct m;
>=20
>=20
> Thanks,
>=20
> Fudong=20
>=20
>=20
>=20
>=20
> On Thursday, August 9, 2012 2:16:52 PM UTC-7, Michael Hunger wrote:
> I don't really understand your question?=20
>=20
> Does the second query not work for you?=20
>=20
> Michael=20
>=20
> Am 09.08.2012 um 21:48 schrieb fli.rpx:=20
>=20
> > I am trying to write a query to return nodes with two relationships.=20=

> >=20
> > One way to do it: start a=3Dnode(1405) match a-[:A*1..3]->m, =
a-[:B*1..3]->m return distinct m.name;=20
> >=20
> > This will work, but takes much a lot more time than a query like =
this: start a=3Dnode(1405) match a-[:A|B*1..3] return distinct m;=20
> >=20
> > I think maybe neo4j does two traversals in the first query.=20
> yes it does and their results span up a cross product which is =
filtered down by distinct again.=20
> >=20
> > Is it possible to do a query to get nodes with both relationships =
like the second query?=20
>=20


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<html><body style=3D"word-wrap: break-word; -webkit-nbsp-mode: space; =
-webkit-line-break: after-white-space; ">you can write the first one =
like this:<br><br><blockquote type=3D"cite">start a=3Dnode(1405) match =
a-[:A*1..3]-&gt;m<br></blockquote><blockquote type=3D"cite"><font =
class=3D"Apple-style-span" =
color=3D"#000000">where&nbsp;</font>a-[:B*1..3]-&gt;m</blockquote><blockqu=
ote type=3D"cite">return distinct&nbsp;<a =
href=3D"http://m.name">m.name</a><br></blockquote><div><br></div>Please =
remember that distinct is an expensive operation (in term of =
non-lazyness and memory consumption) as it has to fetch all data and =
merge the results in memory.<div><br></div><div>You may at least try to =
run it without distinct to see the =
difference.</div><div><br></div><div>Michael<br><div><br>Am 10.08.2012 =
um 18:59 schrieb fli.rpx:<br><br><blockquote type=3D"cite">Hi =
Michael,<br><br>Thanks for your reply.<br><br>There are differences =
between the two queries. The first query, "start a=3Dnode(1405) match =
a-[:A*1..3]-&gt;m, a-[:B*1..3]-&gt;m return distinct&nbsp;<a =
href=3D"http://m.name">m.name</a>", has AND relationship. Its result set =
will be the intersection of two traversals.<br><br>The second query, =
"start a=3Dnode(1405) match a-[:A|B*1..3] return distinct m;&nbsp;", is =
OR relationship.&nbsp;<br><br>These two queries have performance =
difference. The first one will take two traversals then compare. The =
second one only has one traversal.<br><br>I am wondering if I can write =
the first query in a way of the second query. Something like =
this?:<br><br>start a=3Dnode(1405) match a-[:A&amp;B*1..3] return =
distinct m;<br><br><br>Thanks,<br><br>Fudong&nbsp;<br><br><br><br><br>On =
Thursday, August 9, 2012 2:16:52 PM UTC-7, Michael Hunger wrote:<br>I =
don't really understand your question?&nbsp;<br><br>Does the second =
query not work for you?&nbsp;<br><br>Michael&nbsp;<br><br>Am 09.08.2012 =
um 21:48 schrieb fli.rpx:&nbsp;<br><br>&gt; I am trying to write a query =
to return nodes with two relationships.&nbsp;<br>&gt;&nbsp;<br>&gt; One =
way to do it: start a=3Dnode(1405) match a-[:A*1..3]-&gt;m, =
a-[:B*1..3]-&gt;m return =
distinct&nbsp;m.name;&nbsp;<br>&gt;&nbsp;<br>&gt; This will work, but =
takes much a lot more time than a query like this: start a=3Dnode(1405) =
match a-[:A|B*1..3] return distinct m;&nbsp;<br>&gt;&nbsp;<br>&gt; I =
think maybe neo4j does two traversals in the first query.&nbsp;<br>yes =
it does and their results span up a cross product which is filtered down =
by distinct again.&nbsp;<br>&gt;&nbsp;<br>&gt; Is it possible to do a =
query to get nodes with both relationships like the second =
query?&nbsp;<br><br></blockquote><br></div></div></body></html>=

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