As part of a larger query, I am trying to select PRODUCTs that have a relationship to more than one SKU. I subsequently want to delete these relationships and perform other modifications not included here for the sake of simplicity.
I was surprised to find that while the following query returns a single node:
MATCH (p:PRODUCT)-[p_s_rel:PRODUCT_SKU]->(s:SKU)
WITH s, p, count(s) AS sCount
WHERE sCount> 1 AND id(s) IN [9220, 9201]
RETURN s
adding the relationships p_s_rel in the WITH clause changes the result and returns no nodes:
MATCH (p:PRODUCT)-[p_s_rel:PRODUCT_SKU]->(s:SKU)
WITH s, p, p_s_rel, count(s) AS sCount
WHERE sCount> 1 AND id(s) IN [9220, 9201]
RETURN s
Based on the documentation for WITH, I didn't expect this behavior. Is it not possible to specify relationship identifiers in the WITH clause?
If that's the case, how can I delete the p_s_rel relationships in my example?
Edit: This is on version 2.1.5
Are you sure that the first query is doing what you think that it is? In your query if sCount > 1 then I think that means you have more than one relationship between p and some SKU (i.e multiple relationships between the same nodes). If your WITH statement were WITH p, count(s) AS sCount then you would be matching a single Product with multiple SKUs.
By executing WITH s, p, count(s) AS sCount you are saying carry the currently matched SKU, the currently matched PRODUCT and the count of PRODUCT_SKU relationships between them, whereas WITH p, count(s) AS sCount would be taken to mean carry the currently matched PRODUCT and the count of all PRODUCT_SKU relationships that it has.
In your second query, by including p_s_rel in your WITH clause you will be propagating only a single row with each result - as there will only ever be one distinct p_s_rel to carry forward in each match (a relationship only has one start node, and one end node). This means that sCount will never be greater than 1 - hence your empty resultset. i.e you are saying carry the currently matched SKU, the currently matched PRODUCT, the current realtionship between those two nodes and the count of the end nodes on that relationship (1).
If you want to carry the relationships forward you could use COLLECT, or, as you are going to be restricting your resultset by SKU, maybe you would be better off matching those first:
MATCH (s1:SKU), (s2:SKU)
WHERE ID(s1) = 9220 AND ID(s2) = 9201
WITH s1, s2
MATCH (s1)<-[r1:PRODUCT_SKU]-(p:PRODUCT)-[r2:PRODUCT_SKU]->(s2)
DELETE r1, r2,
RETURN p
Related
I am trying to return a set of a node from 2 sessions with a condition that returned node should not be present in another session (third session). I am using the following code but it is not working as intended.
MATCH (:Session {session_id: 'abc3'})-[:HAS_PRODUCT]->(p:Product)
UNWIND ['abc1', 'abc2'] as session_id
MATCH (target:Session {session_id: session_id})-[r:HAS_PRODUCT]->(product:Product)
where p<>product
WITH distinct product.products_id as products_id, r
RETURN products_id, count(r) as score
ORDER BY score desc
This query was supposed to return all nodes present in abc1 & abc2 but not in abc3. This query is not excluding all products present in abc3. Is there any way I can get it working?
UPDATE 1:
I tried to simplify it without UNWIND as this
match (:Session {session_id: 'abc3'})-[:HAS_PRODUCT]->(p:Product)
MATCH (target:Session {session_id: 'abc1'})-[r:HAS_PRODUCT]->(product:Product)
where product <> p
WITH distinct product.products_id as products_id
RETURN products_id
Even this is also not working. It is returning all items present in abc1 without removing those which are already in abc3. Seems like where product <> p is not working correctly.
I would suggest it would be best to check if the nodes are in a list, and to prove out the approach, start with a very simple example.
Here is a simple cypher showing one way to do it. This approach can then be extended into the complex query,
// get first two product IDs as a list
MATCH (p:Product)
WITH p LIMIT 2
WITH COLLECT(ID(p)) as list
RETURN list
// now show two more product IDs which not in that list
MATCH (p:Product)
WITH p LIMIT 2
WITH COLLECT(ID(p)) as list
MATCH (p2:Product)
WHERE NOT ID(p2) in list
RETURN ID(p2) LIMIT 2
Note: I'm using the ID() of the nodes instead of the entire node, same dbhits but may be more performant...
I have developed a query which, by trial and error, appears to find all of the duplicated relationships in a Neo4j DB. I want delete all but one of these relationships but I'm concerned that I have not thought of problematic cases that could result in data deletion.
So, does this query delete all but one of a duplicated relationship?
MATCH (a)-->(b)<--(a) # identify where the duplication is present
WITH DISTINCT a, b
MATCH (a)-[r]->(b) # get all duplicated paths themselves
WITH a, b, collect(r)[1..] as rs # remove the first instance from the list
UNWIND rs as r
DELETE r
If I replace the UNWIND rs as r; DELETE r with WITH a, b, count(rs) as cnt RETURN cnt it seems to return the unnecessary relationships.
I'm still relucant to put this somewhere to be used by others, though....
Thanks
First of all, let me (strictly) define the term: "duplicate relationships". Two relationships are duplicates if they:
Connect the same pair of nodes (call them a and b)
Have the same relationship type
Have exactly the same set of properties (both names and values)
Have the same directionality between a and b (iff directionality is significant for use case)
Your query only considers #1 and #4, so it generally could delete non-duplicate relationships as well.
Here is a query that will take all of the above into consideration (assuming #4 should be included):
MATCH (a)-[r1]->(b)<-[r2]-(a)
WHERE TYPE(r1) = TYPE(r2) AND PROPERTIES(r1) = PROPERTIES(r2)
WITH a, b, apoc.coll.union(COLLECT(r1), COLLECT(r2))[1..] AS rs
UNWIND rs as r
DELETE r
Aggregating functions (like COLLECT) use non-aggregated terms as grouping keys, so there is no need for the query to perform a separate redundant DISTINCT a,b test.
The APOC function apoc.coll.union returns the distinct union of its 2 input lists.
match(m:master_node:Application)-[r]-(k:master_node:Server)-[r1]-(n:master_node)
where (m.name contains '' and (n:master_node:DeploymentUnit or n:master_node:Schema))
return distinct m.name,n.name
Hi,I am trying to get total number of records for the above query.How I change the query using count function to get the record count directly.
Thanks in advance
The following query uses the aggregating funtion COUNT. Distinct pairs of m.name, n.name values are used as the "grouping keys".
MATCH (m:master_node:Application)--(:master_node:Server)--(n:master_node)
WHERE EXISTS(m.name) AND (n:DeploymentUnit OR n:Schema)
RETURN m.name, n.name, COUNT(*) AS cnt
I assume that m.name contains '' in your query was an attempt to test for the existence of m.name. This query uses the EXISTS() function to test that more efficiently.
[UPDATE]
To determine the number of distinct n and m pairs in the DB (instead of the number of times each pair appears in the DB):
MATCH (m:master_node:Application)--(:master_node:Server)--(n:master_node)
WHERE EXISTS(m.name) AND (n:DeploymentUnit OR n:Schema)
WITH DISTINCT m.name AS n1, n.name AS n2
RETURN COUNT(*) AS cnt
Some things to consider for speeding up the query even further:
Remove unnecessary label tests from the MATCH pattern. For example, can we omit the master_node label test from any nodes? In fact, can we omit all label testing for any nodes without affecting the validity of the result? (You will likely need a label on at least one node, though, to avoid scanning all nodes when kicking off the query.)
Can you add a direction to each relationship (to avoid having to traverse relationships in both directions)?
Specify the relationship types in the MATCH pattern. This will filter out unwanted paths earlier. Once you do so, you may also be able to remove some node labels from the pattern as long as you can still get the same result.
Use the PROFILE clause to evaluate the number of DB hits needed by different Cypher queries.
You can find examples of how to use count in the Neo4j docs here
In your case the first example where:
count(*)
Is used to return a count of each returned item should work.
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.
I have a scenario where I have more than 2 random nodes.
I need to get all possible paths connecting all three nodes. I do not know the direction of relation and the relationship type.
Example : I have in the graph database with three nodes person->Purchase->Product.
I need to get the path connecting these three nodes. But I do not know the order in which I need to query, for example if I give the query as person-Product-Purchase, it will return no rows as the order is incorrect.
So in this case how should I frame the query?
In a nutshell I need to find the path between more than two nodes where the match clause may be mentioned in what ever order the user knows.
You could list all of the nodes in multiple bound identifiers in the start, and then your match would find the ones that match, in any order. And you could do this for N items, if needed. For example, here is a query for 3 items:
start a=node:node_auto_index('name:(person product purchase)'),
b=node:node_auto_index('name:(person product purchase)'),
c=node:node_auto_index('name:(person product purchase)')
match p=a-->b-->c
return p;
http://console.neo4j.org/r/tbwu2d
I actually just made a blog post about how start works, which might help:
http://wes.skeweredrook.com/cypher-it-all-starts-with-the-start/
Wouldn't be acceptable to make several queries ? In your case you'd automatically generate 6 queries with all the possible combinations (factorial on the number of variables)
A possible solution would be to first get three sets of nodes (s,m,e). These sets may be the same as in the question (or contain partially or completely different nodes). The sets are important, because starting, middle and end node are not fixed.
Here is the code for the Matrix example with added nodes.
match (s) where s.name in ["Oracle", "Neo", "Cypher"]
match (m) where m.name in ["Oracle", "Neo", "Cypher"] and s <> m
match (e) where e.name in ["Oracle", "Neo", "Cypher"] and s <> e and m <> e
match rel=(s)-[r1*1..]-(m)-[r2*1..]-(e)
return s, r1, m, r2, e, rel;
The additional where clause makes sure the same node is not used twice in one result row.
The relations are matched with one or more edges (*1..) or hops between the nodes s and m or m and e respectively and disregarding the directions.
Note that cypher 3 syntax is used here.