What about this approach ?
WITH 'A' as name
MATCH p=((n:Root)-[:CHILD*]->(m {name:name}))
WHERE SINGLE(n in TAIL(nodes(p)) WHERE n.name=name)
RETURN m
which wil return only the first node with name=‘A’ , if any.
> On 27 Dec 2016, at 11:58, Behrang Saeedzadeh <[email protected]> wrote:
>
> This might work, but my dataset has about 2m nodes and it is very slow as it
> results a cartesian join:
>
> MATCH (n), (p)
> WHERE
> n.name = 'A' AND
> p.name = 'A' AND
> NOT (p)-->(n)
> RETURN n
>
>
> On Tuesday, December 27, 2016 at 9:52:08 PM UTC+11, Behrang Saeedzadeh wrote:
> Hi Benoit,
>
> A is the value of the name property for the nodes. So the query is more like
> this:
>
> MATCH (n)
> WHERE
> size ((m)-->(n)) = 0
> AND m.name <http://m.name/> = 'A'
> AND n.name <http://n.name/> = 'A'
> RETURN n
>
> However this won't work because I don't have m in the MATCH clause:
>
> Variable `m` not defined (line 3, column 9 (offset: 24))
> " size ((m)-->(n)) = 0"
> ^
>
>
>
> On Saturday, December 24, 2016 at 9:33:47 PM UTC+11, Benoît Simard wrote:
> Something like that ?
>
> MATCH (n:A)
> WHERE
> size((:A)-->(n)) = 0 AND
> exists((:Root)-[*]->(n))
> RETURN n
>
>
>
> 2016-12-24 10:40 GMT+01:00 Behrang Saeedzadeh <[email protected] <>>:
> Let's pretend we have a tree like this in which each node is labeled with the
> name property:
>
> ROOT
> |
> \---> X --> A --> Y --> A --> Z
> |
> |
> \---> A --> A
> |
> |
> \---> A
>
> I want to return all the nodes named A that don't have an ancestor that is
> also named A. In other words, only the A nodes that are in bold and are
> underlined.
>
> What Cypher query can return these nodes?
>
> Thanks in advance.
>
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