To find a node's neighbors,I use query sentence as below
MATCH (self:Person {id:"13619240353"})-[r*1..2]-(N) return collect(r)
Then I get the result like this
enter image description here
Here r is a list of relations,thus collect(r) is a list of lists,but I expect to
return a list of relations including all the relations in the collect(r),and without the duplicates.How to write the query?
Since with the variable length of the pattern, the named result is an list, then you need to UNWIND it and use a DISTINCT to remove duplicates:
MATCH (self:Person {id:"13619240353"})-[rs*1..2]-(N)
UNWIND rs AS r
RETURN collect(DISTINCT r)
Related
I want to get all the nodes that have specific labels. My code:
match (n) where labels(n)=["Person","Actor","Old"] return n
While there are nodes that satisfy this property I do not get any results.
This is why:
Labels(n) returns an array of strings, but not necessarily in a specific order.
You can try this:
WHERE n:Person AND n:Actor AND n:Old
Or
WHERE ALL(l in [“Person”, “Actor”, “Old”] WHERE l IN labels(n) )
List does not match if the items are not in the same order/sequence. So first of all, list out all labels in your database so you can see how the labels are arranged.
match (n)
return distinct labels(n)
Then you will see which node will have those labels that you look for: ["Person","Actor","Old"].
If you are trying to find nodes where it contains all nodes in that list in any order then this query will work for you.
match (n)
where all(lbl in ["Person","Actor","Old"] where lbl in labels(n))
return n
if you like using APOC functions, here is the query
match (n)
where apoc.coll.isEqualCollection(["Person","Actor","Old"], labels(n))
return n
I find weird that using OPTIONAL MATCH nodes that don’t have the expected relationship are not returned as a single node in path.
OPTIONAL MATCH path = (:Person) -[:LIKES]- (:Movie)
UNWIND nodes(p) as n
UNWIND rels(p) as e
WITH n
WHERE HEAD(LABELS(n)) = “Person”
return COUNT(DISTINCT n)
The number of people returned only includes those who liked a movie. By using OPTIONAL I would have expected all people to be returned.
Is there a workaround to this or am I doing some this wrong in the query?
A better way to go about this would be to match to all :People nodes first, then use the OPTIONAL MATCH to match to movies (or, if you want a collection of the movies they liked, use pattern comprehension).
If you do need to perform an UNWIND on an empty collection without wiping out the row, use a CASE around some condition to use a single-element list rather than the empty list.
MATCH (n:Person) // match all persons
OPTIONAL MATCH p = (n) -[:LIKES]- (m:Movie) // p and m are the optionals
UNWIND CASE WHEN p is null THEN [null] ELSE nodes(p) END as nodes // already have n, using a different variable
UNWIND CASE WHEN p is null THEN [null] ELSE rels(p) END as e // forcing a single element list means UNWIND won't wipe out the row
WITH n
WHERE HEAD(LABELS(n)) = “Person” // not really needed at all, and bad practice, you don't know the order of the labels on a node
return COUNT(DISTINCT n) // if this is really all you need, just keep the first match and the return of the query (without distinct), don't need anything else
Background
Hi all, I am currently trying to write a cypher statement that allows me to find a set of paths on a map from a starting point. I want my search result to always return connecting streets within 5 nodes. Optionally, if there's a nearby hospital, I would like my search pattern to also indicate nearby hospitals.
Main Problem
Because there isn't always a nearby hospital to the current street, sometimes my optional match search pattern comes back as null. Here's the current cypher statement I'm using:
MATCH path=(a:Street {id: 123})-[:CONNECTED_TO*..5]-(b:Street)
OPTIONAL MATCH optionalPath=(b)-[:CONNECTED_TO]->(hospital:Hospital)
WHERE ALL (x IN nodes(path) WHERE (x:Street))
WITH DISTINCT nodes(path) + nodes(optionalPath) as n
UNWIND n as nodes
RETURN DISTINCT nodes;
However, this syntax only works if optionalPath contains nodes. If it doesn't, the statement nodes(path) + nodes(optionalPath) is an operation adding null and I get no records. This is true even the nodes(path) term does contain nodes.
What's the best way to get around this problem?
You can use COALESCE to replace a NULL with some other value. For example:
MATCH path=(:Street {id: 123})-[:CONNECTED_TO*..5]-(b:Street)
WHERE ALL (x IN nodes(path) WHERE x:Street)
OPTIONAL MATCH optionalPath=(b)-[:CONNECTED_TO]->(hospital:Hospital)
WITH nodes(path) + COALESCE(nodes(optionalPath), []) as n
UNWIND n as nodes
RETURN DISTINCT nodes;
I have also made a few other improvements:
The WHERE clause was moved up right after the first MATCH. This eliminates the unwanted path values immediately. Your original query would get all path values (even unwanted ones) and always the perform the second MATCH query, and only eliminate unwanted paths afterwards. (But, it is actually not clear if you even need the WHERE clause at all; for example, if the CONNECTED_TO relationship is only used between Street nodes.)
The DISTINCT in your WITH clause would have prevented duplicate n collections, but the collections internally could have had duplicate paths. This was probably not what you wanted.
It seems you don't really want the path, just all the street nodes within 5 steps, plus any connected hospitals. So I would simplify your query to just that, and then condense the 3 columns down to 1.
MATCH (a:Street {id: 123})-[:CONNECTED_TO*..5]-(b:Street)
OPTIONAL MATCH (b)-[:CONNECTED_TO]->(hospital:Hospital)
WITH collect(a) + collect(b) + collect(hospital) as n
UNWIND n as nodez
RETURN DISTINCT nodez;
If Streets can be indirectly connected (hospital in between), Than I'd adjust like this
MATCH (a:Street {id: 123})-[:CONNECTED_TO]-(b:Street)
WITH a as nodez, b as a
MATCH (a)-[:CONNECTED_TO]-(b:Street)
WITH nodez+collect(b) as nodez, b as a
MATCH (a)-[:CONNECTED_TO]-(b:Street)
WITH nodez+collect(b) as nodez, b as a
MATCH (a)-[:CONNECTED_TO]-(b:Street)
WITH nodez+collect(b) as nodez, b as a
MATCH (a)-[:CONNECTED_TO]-(b:Street)
WITH nodez+collect(b) as nodez, b as a
OPTIONAL MATCH (b)-[:CONNECTED_TO]->(hospital:Hospital)
WITH nodez + collect(hospital) as n
UNWIND n as nodez
RETURN DISTINCT nodez;
It's a bit more verbose, but just says exactly what you want (and also adds the start node to the hospital check list)
So as a complication to this question, I basically want to do
MATCH (n:TEST) OPTIONAL MATCH (n)-[r]->() RETURN DISTINCT n, r
And I want to return n and r as one column with no repeat values. However, running
MATCH (n:TEST) OPTIONAL MATCH (n)-[r]->() UNWIND n+r AS x RETURN DISTINCT x
gives a "Type mismatch: expected List but was Relationship (line 1, column 47)" error. And this query
MATCH (n:TEST) RETURN DISTINCT n UNION MATCH ()-[n]->() RETURN DISTINCT n
Puts nodes and relationships in the same column, but the context from the first match is lost in the second half.
So how can I return all matched nodes and relationships as one minimal list?
UPDATE:
This is the final modified version of the answer query I am using
MATCH (n:TEST)
OPTIONAL MATCH (n)-[r]->()
RETURN n {.*, rels:collect(r {properties:properties(r), id:id(r), type:type(r), startNode:id(startNode(r)), endNode:id(endNode(r))})} as n
There are a couple ways to handle this, depending on if you want to hold these within lists, or within maps, or if you want a map projection of a node to include its relationships.
If you're using Neo4j 3.1 or newer, then map projection is probably the easiest approach. Using this, we can output the properties of a node and include its relationships as a collected property:
MATCH (n:TEST)
OPTIONAL MATCH (n)-[r]->()
RETURN n {.*, rels:collect(r)} as n
Here's what you might do if you wanted each row to be its own pairing of a node and a single one of its relationships as a list:
...
RETURN [n, r] as pair
And as a map:
...
RETURN {node:n, rel:r} as pair
EDIT
As far as returning more data from each relationship, if you check the Code results tab, you'll see that the id, relationship type, and start and end node ids are included, and accessible from your back-end code.
However, if you want to explicitly return this data, then we just need to include it in the query, using another map projection for each relationship:
MATCH (n:TEST)
OPTIONAL MATCH (n)-[r]->()
RETURN n {.*, rels:collect(r {.*, id:id(r), type:type(r), startNode:startNode(r), endNode:endNode(r)})} as n
I have a graph in that contains two types of nodes (objects and pieces) and two types of links (similarTo and contains). Some pieces are made of the pieces.
I would like to extract the path to each piece starting from a set of objects.
MATCH (o:Object)
WITH o
OPTIONAL MATCH path = (p:Piece) <-[:contains*]- (o) -[:similarTo]- (:Object)
RETURN path
The above query only returns part of the pieces. In the returned graph, some objects do not directly connect to any pieces, the latter are not returned, although they actually do!
I can change the query to:
MATCH (o:Object) -[:contains*]-> (p:Piece)
OPTIONAL MATCH (o) –[:similarTo]- (:Object)
However, I did not manage to return the whole path for that query, which I need to return collection of nodes and links with:
WITH rels(path) as relations , nodes(path) as nodes
UNWIND relations as r unwind nodes as n
RETURN {nodes: collect(distinct n), links: collect(distinct {source: id(startNode(r)), target: id(endNode(r))})}
I'd be grateful to any recommendation.
Would something like this do the trick ?
I created a small graph representing objects and pieces here : http://console.neo4j.org/r/abztz4
Execute distinct queries with UNION ALL
Here you'll combine the two use cases in one set of paths :
MATCH (o:Object)
WITH o
OPTIONAL MATCH p=(o)-[:CONTAINS]->(piece)
RETURN p
UNION ALL
MATCH (o:Object)
WITH o
OPTIONAL MATCH p=(o)-[:SIMILAR_TO]-()-[:CONTAINS]->(piece)
RETURN p