I have a linked list that is modelled in a circular fashion like so:
(u:User)
-[:LINK]->(a:NODELINK {linkId: 'aa'})
-[:LINK]->(b:NODELINK {linkId: 'bb'})
-[:LINK]->(c:NODELINK {linkId: 'cc'})
-[:LINK]->(d:NODELINK {linkId: 'dd'})
-[:LINK]->(u)
When I query it starting at node (b:NODELINK {linkId: 'bb'}) I would like to match all nodes until I get to the end/start of the list. (The User node)
When I run the following query:
MATCH (u:USER)
WITH u
MATCH (nl:NODELINK)-[:LINK*]->(m)
WHERE nl.linkId = 'bb' AND m <> u
RETURN m
It returns me the following nodes
(3:NODELINK {linkId:'cc'})
(4:NODELINK {linkId:'dd'})
(1:NODELINK {linkId:'aa'})
(2:NODELINK {linkId:'bb'})
As you can see my query wraps around and starts returning nodes from the start of the list. I would like to terminate at the User node and only return the nodes before the end of the list
(3:NODELINK {linkId:'cc'})
(4:NODELINK {linkId:'dd'})
I have created a graph that demonstrates the issue here
How do I query so as to start at the node link of interest and returns all nodes before it reaches the User node?
Your WHERE clause limits the first and last node in the path, but not the intermediate ones, so yes, the query can match paths that include the user node.
Instead of matching many paths and returning only the end nodes, try matching one path from the 'cc' node to the user node, and return all nodes in that path (except the last one if you don't want the user node).
MATCH path=(nl:NODELINK {linkId:'cc'})-[:LINK*]->(u:USER)
RETURN nodes(path)
or
RETURN nodes(path)[..-1]
if you want to return all the nodes except the last one.
Try below query. I think you should bring the NODELINK where you want to begin traversing in the first MATCH clause as it will pin point your graph where to start instead of just matching user node on USER label.
MATCH (nl:NODELINK)
WHERE nl.linkId = 'aa'
WITH nl
MATCH p=(nl)-[:LINK*1..]->(m:USER)
RETURN filter(t IN nodes(p)
WHERE t <> nl) AS nodes
Related
I have a graph of nodes with a relationship NEXT with 2 properties sequence (s) and position (p). For example:
N1-[NEXT{s:1, p:2}]-> N2-[NEXT{s:1, p:3}]-> N3-[NEXT{s:1, p:4}]-> N4
A node N might have multiple outgoing Next relationships with different property values.
Given a list of node names, e.g. [N2,N3,N4] representing a sequential path, I want to check if the graph contains the nodes and that the nodes are connected with relationship Next in order.
For example, if the list contains [N2,N3,N4], then check if there is a relationship Next between nodes N2,N3 and between N3,N4.
In addition, I want to make sure that the nodes are part of the same sequence, thus the property s is the same for each relationship Next. To ensure that the order maintained, I need to verify if the property p is incremental. Meaning, the value of p in the relationship between N2 -> N3 is 3 and the value p between N3->N4 is (3+1) = 4 and so on.
I tried using APOC to retrieve the possible paths from an initial node N using python (library: neo4jrestclient) and then process the paths manually to check if a sequence exists using the following query:
q = "MATCH (n:Node) WHERE n.name = 'N' CALL apoc.path.expandConfig(n {relationshipFilter:'NEXT>', maxLevel:4}) YIELD path RETURN path"
results = db.query(q,data_contents=True)
However, running the query took some time that I eventually stopped the query. Any ideas?
This one is a bit tough.
First, pre-match to the nodes in the path. We can use the collected nodes here to be a whitelist for nodes in the path
Assuming the start node is included in the list, a query might go like:
UNWIND $names as name
MATCH (n:Node {name:name})
WITH collect(n) as nodes
WITH nodes, nodes[0] as start, tail(nodes) as tail, size(nodes)-1 as depth
CALL apoc.path.expandConfig(start, {whitelistNodes:nodes, minLevel:depth, maxLevel:depth, relationshipFilter:'NEXT>'}) YIELD path
WHERE all(index in range(0, size(nodes)-1) WHERE nodes[index] = nodes(path)[index])
// we now have only paths with the given nodes in order
WITH path, relationships(path)[0].s as sequence
WHERE all(rel in tail(relationships(path)) WHERE rel.s = sequence)
// now each path only has relationships of common sequence
WITH path, apoc.coll.pairsMin([rel in relationships(path) | rel.p]) as pairs
WHERE all(pair in pairs WHERE pair[0] + 1 = pair[1])
RETURN path
Take the above image as an example. Using Cypher, how would I match all of the nodes except for the longest chain and the central node? I.e. all nodes within exactly one hop of the central node whilst excluding the central node (all nodes and edges except 3 nodes and 2 edges).
I have tried the following:
MATCH (n:Node) WHERE n.id = "123" MATCH path = (m)-[*1..1]->(n) RETURN m
This very nearly works, however it still returns the central node (i.e. node n). How would I exclude this node from my query result?
[UPDATED]
This will return all distinct nodes directly connected to the specified node, and explicitly prevents the specified node from being returned (in case it has a relationship to itself):
MATCH (n:Node)--(m)
WHERE n.id = "123" AND n <> m
RETURN DISTINCT m;
Ideally I would have liked to match the nodes as mentioned in my question and delete them. However, as I have not found a way to do so an inverse approach can be utilised whereby all nodes but those as mentioned in the question are matched instead. Thereby effectively excluding (but not deleting) the unwanted nodes.
This can be achieved using this query:
MATCH (n:Node) WHERE n.id = "123" MATCH path = (m)-[*2..]->(n) RETURN path
This returns the central node and all paths to that node that have a "length" greater than or equal to 2.
According to this post I tried to map all related entities in a list.
I used the same query into the post with a condition to return a list of User but it returns duplicate object
MATCH (user:User) WHERE <complex conditions>
WITH user, calculatedValue
MATCH p=(user)-[r*0..1]-() RETURN user, calculatedValue, nodes(p), rels(p)
Is it a bug? I'm using SDN 4.2.4.RELEASE with neo4j 3.2.1
Not a bug.
Keep in mind a MATCH in Neo4j will find all occurrences of a given pattern. Let's look at your last MATCH:
MATCH p=(user)-[r*0..1]-()
Because you have a variable match of *0..1, this will always return at least one row with just the user itself (with rels(p) empty and nodes(p) containing only the user), and then you'll get a row for every connected node (user will always be present on that row, and in the nodes(p) collection, along with the other connected node).
In the end, when you have a single user node and n directly connected nodes, you will get n + 1 rows. You can run the query in the Neo4j browser, looking at the table results, to confirm.
A better match might be something like:
...
OPTIONAL MATCH (user)-[r]-(b)
RETURN user, calculatedValue, collect(r) as rels, collect(b) as connectedNodes
Because we aggregate on all relationships and connected nodes (rather than just the relationships and nodes for each path), you'll get a single row result per user node.
I'm using Neo4j to try to find any node that is not connected to a specific node "a". The query that I have so far is:
MATCH p = shortestPath((a:Node {id:"123"})-[*]-(b:Node))
WHERE p IS NULL
RETURN b.id as b
So it tries to find the shortest path between a and b. If it doesn't find a path, then it returns that node's id. However, this causes my query to run for a few minutes then crashes when it runs out of memory. I was wondering if this method would even work, and if there is a more efficient way? Any help would be greatly appreciated!
edit:
MATCH (a:Node {id:"123"})-[*]-(b:Node),
(c:Node)
WITH collect(b) as col, a, b, c
WHERE a <> b AND NOT c IN col
RETURN c.id
So col (collect(b)) contains every node connected to a, therefore if c is not in col then c is not connected to a?
For one, you're giving this MATCH an impossible predicate to fulfill, so it will never find the shortest path.
WHERE clauses are associated with MATCH, OPTIONAL MATCH, and WITH clauses, so your query is asking for the shortest path where the path doesn't exist. That will never return anything.
Also, the shortestPath will start at the node you DON'T want to be connected, so this has no way of finding the nodes that aren't connected to it.
Probably the easiest way to approach this is to MATCH to all nodes connected to your node in question, then MATCH to all :Nodes checking for those that aren't in the connected set. That means you won't have to do a shortestPath from every single node in the db, just a membership check in a collection.
You'll need APOC Procedures for this, as it has the fastest means of matching to nodes within a subgraph.
MATCH (a:Node {id:"123"})
CALL apoc.path.subgraphNodes(a, {}) YIELD node
WITH collect(node) as subgraph
MATCH (b:Node)
WHERE NOT b in subgraph
RETURN b.id as b
EDIT
Your edited query is likely to blow up, that's going to generate a huge result set (the query will build a result set of every node reachable from your start node by a unique path in a cartesian product with every :Node).
Instead, go step by step, collect the distinct nodes (because otherwise you'll get multiples of the same nodes that can be reached via different paths), and then only after you have your collection should you start your match for nodes that aren't in the list.
MATCH (:Node {id:"123"})-[*0..]-(b:Node)
WITH collect(DISTINCT b) as col
MATCH (a:Node)
WHERE NOT a IN col
RETURN a.id
I'm trying to find the number of nodes of a certain kind in my database that are connected to more than one other node of another kind. In my case, it's place nodes connected to several name nodes. I have a query that works:
MATCH rels=(p:Place)-[c:Called]->(n:Name)
WITH p,count(n) as counts
WHERE counts > 1
RETURN p;`
However, that only returns the place nodes, and ideally I'd like it to return all the nodes and edges involved. I've found a question on returning variables from before the WITH, but if I include any of the other variables I've defined, the query returns no responses, i.e. this query returns nothing:
MATCH rels=(p:Place)-[c:Called]->(n:Name)
WITH p, count(n) as counts, rels
WHERE counts > 1
RETURN p;
I don't know how to return the information that I want without changing the results of the query. Any help would be much appreciated
The reason your second query returns nothing is because its WITH clause specifies as aggregation "grouping keys" both p and rels. Since each rels path has only a single n value, counts would always be 1.
Something like this might work for you:
MATCH path=(p:Place)-[:Called]->(:Name)
WITH p, COLLECT(path) as paths
WHERE SIZE(paths) > 1
RETURN p, paths;
This returns each matching Place node and all its paths.
Try this:
MATCH (p:Place)-[c:Called]->(n:Name)
WHERE size((p)-[:Called]->(:Name)) > 1
WITH p,count(n) as counts, collect(n) AS names, collect(c) AS calls
RETURN p, names, calls, counts ORDER BY counts DESC;
This query makes use of Cypher's collect() function to create lists of the names and called relationships for each place that has more than Called relationship with a Name node.