What's the Cypher equivalent to "find all nodes that have a property of name with value of X that don't have any ancestors with name = X"?

[Behrang Saeedzadeh]

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.

[Benoît Simard]

Something like that ?

MATCH (n:A)

WHERE 

  size((:A)-->(n)) = 0 AND

 exists((:Root)-[*]->(n))

RETURN n

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[Behrang Saeedzadeh]

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 = 'A'

 AND 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 <behr...@gmail.com>:

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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[Behrang Saeedzadeh]

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

[Tom Zeppenfeldt]

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.

[Behrang Saeedzadeh]

For some reason this is returning such nodes too:

 /ROOT/N1/A/N2/A

I will upload the database sometime later. I think that should make it easier to test the queries.

[Max De Marzi Jr.]

I think Tom almost had it...either:

WITH 'A' as name 

          MATCH p=((n:Root)-[:CHILD*]->(m {name:name})) 

          WHERE SINGLE(n in nodes(p) WHERE n.name=name) 

          RETURN m

or maybe:

WITH 'A' as name 

          MATCH p=((n:Root)-[:CHILD*]->(m {name:name})) 

          WHERE NONE(n in TAIL(nodes(p)) WHERE n.name=name) 

          RETURN m

[Behrang Saeedzadeh]

Thanks. The first query worked. I have attached a sample DB for those who are interested.