Return nodes in shortestPath - neo4j

Given following database schema:
I would like to return the movie that connects the actor Bruce Wilis to the director Oliver Stone using the shortestPath() function.
My attempt:
match p = shortestPath((d:Director{director_name:'Oliver Stone'})-[*]- (a:Actor{actor_name:'Bruce Willis'}))
with nodes(p) as n
where type(n) = 'Movie'
return n

The problem in your attempted query is that nodes(p) is a collection of nodes. Trying to use type() on a collection of nodes won't work. Also, type() is only used to get the type of a relationship. Nodes don't have types, they have labels.
Also, you don't know if there is just a single movie node, or multiple movie nodes, in the shortest path between your starting nodes, so the best approach is to filter the collection so you only keep movie nodes in the path:
match p = shortestPath((d:Director{director_name:'Oliver Stone'})-[*]- (a:Actor{actor_name:'Bruce Willis'}))
return [n in nodes(p) where n:Movie] as movies
This uses list comprehension to perform the filtering of the collection. Alternately, you can use the filter() function:
return filter(n in nodes(p) where n:Movie) as movies

Related

Cypher - given relationship, get the nodes

If I do the query
MATCH (:Label1 {prop1: "start node"}) -[relationships*1..10]-> ()
UNWIND relationships as relationship
RETURN DISTINCT relationship
How do I get nodes for each of acquired relationship to get result in format:
╒════════╤════════╤═══════╕
│"from" │"type" │"to" │
╞════════╪════════╪═══════╡
├────────┼────────┼───────┤
└────────┴────────┴───────┘
Is there a function such as type(r) but for getting nodes from relationship?
RomanMitasov and ray have working answers above.
I don't think they quite get at what you want to do though, because you're basically returning every relationship in the graph in a sort of inefficient way. I say that because without a start or end position, specifying a path length of 1-10 doesn't do anything.
For example:
CREATE (r1:Temp)-[:TEMP_REL]->(r2:Temp)-[:TEMP_REL]->(r3:Temp)
Now we have a graph with 3 Temp nodes with 2 relationships: from r1 to r2, from r2 to r3.
Run your query on these nodes:
MATCH (:Temp)-[rels*1..10]->(:Temp)
UNWIND rels as rel
RETURN startNode(rel), type(rel), endNode(rel)
And you'll see you get four rows. Which is not what you want because there are only two distinct relationships.
You could modify that to return only distinct values, but you're still over-searching the graph.
To get an idea of what relationships are in the graph and what they connect, I use a query like:
MMATCH (n)-[r]->(m)
RETURN labels(n), type(r), labels(m), count(r)
The downside of that, of course, is that it can take a while to run if you have a very large graph.
If you just want to see the structure of your graph:
CALL db.schema.visualization()
Best wishes and happy graphing! :)
Yes, such functions do exist!
startNode(r) to get the start node from relationship r
endNode(r) to get the end node
Here's the final query:
MATCH () -[relationships*1..10]-> ()
UNWIND relationships as r
RETURN startNode(r) as from, type(r) as type, endNode(r) as to

How Many Nodes Are Involved in a Match

How can I know how many nodes and edges are involved in a MATCH? Is there another way besides Explain / Profile Match?
If you mean how many nodes are matched in a path, such as a variable-length path, then you can assign a path variable for this:
MATCH p = (k:Person {name:'Keanu Reeves'})-[*..8]-(t:Person {name:'Tom Hanks'})
WITH p LIMIT 1
RETURN p, length(p) as pathLength, length(p) + 1 as numberOfNodesInPath
You can also use nodes(p) and relationships(p) to get the collection of nodes and relationships that make up the path, and you can use size() on those collections to get their size.
There exists the COUNT() function of Cypher that allows you to count the number of elements. As for example in this query:
MATCH (n)
RETURN COUNT(n);
This query will count all nodes in your database.
You can find more information in the cypher manual, under the aggregating functions. Check it out.
The following Cypher snippet should return the number of distinct nodes and relationships found by any given MATCH clause. Just replace <your code here> with your MATCH pattern.
MATCH <your code here>
WITH COLLECT(NODES(p)) AS ns, SUM(SIZE(RELATIONSHIPS(p))) AS relCount
UNWIND ns AS nodeList
UNWIND nodeList AS node
RETURN COUNT(DISTINCT node) AS nodeCount, relCount;

Return a graph based on the number of relationships between nodes

I am experimenting with creating multiple relationships between nodes to represent the importance between two given nodes.
For example, I want to know what 'genre' of reading material is most important to Joe.
I want a way to match the Joe node to genre nodes only if there is some number or greater relationships between them.
So, if I want matches with 3 or more relationships, I should get a graph with Joe --> Fantasy
I know I can get this when both endpoints are defined:
MATCH (p:PERSON {name:'Joe'})-[r]->(g: GENRE {name:'Fantasy'})
RETURN count(r)
What I want is something like:
MATCH p = (p:PERSON {name:'Joe'})-[r]->()
WHERE *pair_relationship_count*(r) >= 3
RETURN p
This is my proposition:
MATCH path = (p:PERSON {name:'Joe'})-[r]->()
WITH collect(path) as paths, collect(r) as pair_count
WITH paths WHERE size(pair_count) >= 3
UNWIND paths as path
RETURN path
But maybe it is more efficient to have one relationship with an internal count property on one relationship for each couple of nodes.
First, I think you can achieve your goal using WITH clause:
MATCH path = (:PERSON {name:'Joe'})-[r]->(:GENRE {name:'Fantasy'})
WITH path, count(r) as count
WHERE count > 3
RETURN path
But using one relationship for each read "event" seems to be a bad approach. Maybe you should use an integer property in the relationship, then increment this property for each "read". This way you can do queries like:
MATCH path = (:PERSON {name:'Joe'})-[r]->(:GENRE {name:'Fantasy'})
WHERE r.count > 3
RETURN path
To get a collection of all READS paths for "Joe" that involve each genre that he has read at least 3 times:
MATCH p = (:PERSON {name:'Joe'})-[:READS]->(genre)
WITH genre, COLLECT(p) AS paths
WHERE SIZE(paths) >= 3
RETURN genre, paths;

cypher to combine nodes and relationships into a single column

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

Cypher Optional Match

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

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