I am new at Neo4j but not to graphs and I have a specific problem I did not manage to solve with Cypher.
With this type of data:
I would like to be able in a single query to follow some incoming and some outgoing flow.
Example:
Starting on "source"
Follow all "A" relationships in the outgoing way
Follow all "B" relationships in the incoming way
My problem is that Cypher only allows one single direction to be specified in the relationship pattern.
So I could do (source)-[:A|:B*]->() or (source)<-[:A|:B*]-().
But I have no possibility to tell Cypher that I want to follow -[:A]-> and <-[:B]-.
By the way, I know that I could do -[:A|:B]- but this won't solve my problem because I don't want to follow -[:B]-> and <-[:A]-.
Thanks in advance for your help :)
Alternatively to #Gabor Szarnyas answer, I think you can achieve your goal using the APOC procedure apoc.path.expand.
Using this sample data set:
CREATE (:Source)-[:A]->()-[:A]->()<-[:B]-()-[:A]->()
And calling apoc.path.expand:
match (source:Source)
call apoc.path.expand(source,"A>|<B","",0,5) yield path
return path
You will get this path as output:
The apoc.path.expand call starts from the source node following -[:A]-> and <-[:B]- relationships.
Remember to install APOC procedures according to the version of Neo4j you are using. Take a look in the version compatibility matrix.
To express this in a single query would require a regular path query, which has been proposed to and accepted to openCypher, but it is not yet implemented.
I see two possible workarounds. I recreated your example with this command with a Source label for the source node:
CREATE (:Source)-[:A]->()-[:A]->()<-[:B]-()-[:A]->()
(1) Insert additional relationships that have the same direction:
MATCH (s)-[:B]->(t)
CREATE (s)<-[:B2]-(t)
And use this relationship for traversal:
MATCH p=(source)-[:A|:B2*]->()
RETURN p
(2) As you mentioned:
By the way, I know that I could do -[:A|:B]- but this won't solve my problem because I don't want to follow -[:B]-> and <-[:A]-.
You could use this approach to first get potential path candidates and manually check the directions of the relationships afterwards. Of course, this is an expensive operation but you only have to calculate it on the candidates, a possibly small data set.
MATCH p=(source:Source)-[:A|:B*]-()
WITH p, nodes(p) AS nodes, relationships(p) AS rels
WHERE all(i IN range(0, size(rels) - 1) WHERE
CASE type(rels[i])
WHEN 'A' THEN startNode(rels[i]) = nodes[i]
ELSE /* B */ startNode(rels[i]) = nodes[i+1]
END)
RETURN p
Let's break down how this works:
We store path candidates in p and use the nodes and relationships functions to extract the lists of nodes/relationships from it.
We define a range of indexes for the relationships (e.g. from 0, 1, 2 if there are 3 relationships).
To determine the direction of relationships, we use the startNode function. For example, if there is a relationship r between nodes n1 to n2, the paths will like <n1, r, n2>. If r was traversed to in the outgoing direction, the startNode(r) will return n1, if it was traverse in the incoming direction, startNode(r) will return n2. The type of the relationship is checked with the type function and a simple CASE expression is used to differentiate between types.
The WHERE clause uses the all predicate function to check whether all :A and :B relationships had the appropriate directions.
Related
I had another thread about this where someone suggested to do
MATCH (p:Person {person_id: '123'})
WHERE ANY(x IN $names WHERE
EXISTS((p)-[:BELONGS]-(:Face)-[:CORRESPONDS]-(:Image)-[:HAS_ACCESS_TO]-(:Dias {group_name: x})))
MATCH path=(p)-[:ASSOCIATED_WITH]-(:Person)
RETURN path
This does what I need it to, returns nodes that fit the criteria without returning the relationships, but now I need to include another param that is a list.
....(:Dias {group_name: x, second_name: y}))
I'm unsure of the syntax.. here's what I tried
WHERE ANY(x IN $names and y IN $names_2 WHERE..
this gives me a syntax error :/
Since the ANY() function can only iterate over a single list, it would be difficult to continue to use that for iteration over 2 lists (but still possible, if you create a single list with all possible x/y combinations) AND also be efficient (since each combination would be tested separately).
However, the new existenial subquery synatx introduced in neo4j 4.0 will be very helpful for this use case (I assume the 2 lists are passed as the parameters names1 and names2):
MATCH (p:Person {person_id: '123'})
WHERE EXISTS {
MATCH (p)-[:BELONGS]-(:Face)-[:CORRESPONDS]-(:Image)-[:HAS_ACCESS_TO]-(d:Dias)
WHERE d.group_name IN $names1 AND d.second_name IN $names2
}
MATCH path=(p)-[:ASSOCIATED_WITH]-(:Person)
RETURN path
By the way, here are some more tips:
If it is possible to specify the direction of each relationship in your query, that would help to speed up the query.
If it is possible to remove any node labels from a (sub)query and still get the same results, that would also be faster. There is an exception, though: if the (sub)query has no variables that are already bound to a value, then you would normally want to specify the node label for the one node that would be used to kick off that (sub)query (you can do a PROFILE to see which node that would be).
I am trying to query using Neo4j.
I would like to print result of obtaining information while AUTO-COMPLETE is ON in Neo4j.
For example, suppose query that creating 3 nodes as shown below.
create (david:Person {name: 'david'}), (mike:Person {name: 'mike'}), (book:Book {title:'book'}), (david)-[:KNOWS]->(mike), (david)-[:WRITE]->(book), (mike)-[:WRITE]->(book)
Here are 2 images:
Auto-complete on
Auto-complete off
Figure is shown after query, and I would like to obtain all relating node’s relationships based on starting node ('book' node).
I used this query as shown below.
match (book:Book)-[r]-(person) return book, r, person
Whether AUTO-COMPLETE is ON or OFF, I expect to obtain all node’s relationships including “David knows Mike”, but system says otherwise.
I studied a lot of Syntax structure at neo4j website, and somehow it is very difficult for me. So, I upload this post to acquire assistance for you.
You have to return all the data that you need yourself explicitly. It would be bad for Neo4j to automatically return all the relationships for a super node with thousands of relationships for example, as it would mean lots of I/O, possibly for nothing.
MATCH (book:Book)-[r]-(person)-[r2]-()
RETURN book, r, person, collect(r2) AS r2
Thanks to InverseFalcon, this is my query that works.
MATCH p = (book:Book)-[r]-(person:Person)
UNWIND nodes(p) as allnodes WITH COLLECT(ID(allnodes)) AS ALLID
MATCH (a)-[r2]-(b)
WHERE ID(a) IN ALLID AND ID(b) IN ALLID
WITH DISTINCT r2
RETURN startNode(r2), r2, endNode(r2)
How can we add a relationship to the query.
Say A-[C01]-B-[C02]-D and A-[C01]-B-[C03]-E
C01 C02 C03 are relationship codes I want to get output
B E
because I want only nodes that can be reached unbroken by C01 or C03
How can I get this result in Cypher?
You may want to clarify, what you're asking for seems like a very simple case of matching. You may want to provide some more info, such as node labels and how you're matching to your start nodes, since without these we have to make things up for example code.
MATCH (a:Thing)
WHERE a.ID = 123
WITH a
MATCH (a)-[:C01|C03*]->(b:Thing)
RETURN b
The key here is specifying multiple relationship types to traverse, using * for multiplicity, so it will match on all nodes that can be reached by any chain of those relationships.
In a graph where the following nodes
A,B,C,D
have a relationship with each nodes successor
(A->B)
and
(B->C)
etc.
How do i make a query that starts with A and gives me all nodes (and relationships) from that and outwards.
I do not know the end node (C).
All i know is to start from A, and traverse the whole connected graph (with conditions on relationship and node type)
I think, you need to use this pattern:
(n)-[*]->(m) - variable length path of any number of relationships from n to m. (see Refcard)
A sample query would be:
MATCH path = (a:A)-[*]->()
RETURN path
Have also a look at the path functions in the refcard to expand your cypher query (I don't know what exact conditions you'll need to apply).
To get all the nodes / relationships starting at a node:
MATCH (a:A {id: "id"})-[r*]-(b)
RETURN a, r, b
This will return all the graphs originating with node A / Label A where id = "id".
One caveat - if this graph is large the query will take a long time to run.
I'm struggling to find a single clean, efficient Cypher query that will let me identify all distinct paths emanating from a start node such that every relationship in the path is of the same type when there are many relationship types.
Here's a simple version of the model:
CREATE (a), (b), (c), (d), (e), (f), (g),
(a)-[:X]->(b)-[:X]->(c)-[:X]->(d)-[:X]->(e),
(a)-[:Y]->(c)-[:Y]->(f)-[:Y]->(g)
In this model (a) has two outgoing relationship types, X and Y. I would like to retrieve all the paths that link nodes along relationship X as well as all the paths that link nodes along relationship Y.
I can do this programmatically outside of cypher by making a series of queries, the first to
retrieve the list of outgoing relationships from the start node, and then a single query (submitted together as a batch) for each relationship. That looks like:
START n=node(1)
MATCH n-[r]->()
RETURN COLLECT(DISTINCT TYPE(r)) as rels;
followed by:
START n=node(1)
MATCH n-[:`reltype_param`*]->()
RETURN p as path;
The above satisfies my need, but requires at minimum 2 round trips to the server (again, assuming I batch together the second set of queries in one transaction).
A single-query approach that works, but is horribly inefficient is the following single Cypher query:
START n=node(1)
MATCH p = n-[r*]->() WHERE
ALL (x in RELATIONSHIPS(p) WHERE TYPE(x) = TYPE(HEAD(RELATIONSHIPS(p))))
RETURN p as path;
That query uses the ALL predicate to filter the relationships along the paths enforcing that each relationship in the path matches the first relationship in the path. This, however, is really just a filter operation on what it essentially a combinatorial explosion of all possible paths --- much less efficient than traversing a relationship of a known, given type first.
I feel like this should be possible with a single Cypher query, but I have not been able to get it right.
Here's a minor optimization, at least non-matching the paths will fail fast:
MATCH n-[r]->()
WITH distinct type(r) AS t
MATCH p = n-[r*]->()
WHERE type(r[-1]) = t // last entry matches
RETURN p AS path
This is probably one of those things that should be in the Java API if you want it to be really performant, though.