I am trying to combine 2 cyphers into one for performance but have not succeeded.
I need to get the count of multiple properties unique to eachother in the same cypher.
EX 1:
Match (n)
RETURN n.foo, count(*) AS count
EX 2:
Match (n)
RETURN n.bar, count(*) AS count
I was hoping I could just run both:
Match (n)
RETURN n.foo, count(*) AS fooCount, n.bar, count(*) AS barCount
But this returns the same count for both as it is finding where they both match. Not what I want.
So was looking for a way to group them to be unique like:
Match (n)
RETURN {n.foo, count(*) AS fooCount}, {n.bar, count(*) AS barCount}
Obviously this is not valid syntax but shows what I am trying to do.
Any assistance on this is of course appreciated.
It's best to do this back to back, all at once isn't a good idea for this kind of query, as aggregation won't work in your favor.
You could try this:
MATCH (n)
WITH n.bar as bar, count(*) AS count
WITH collect({bar:bar, count:count}) as barCounts
MATCH (n)
WITH barCounts, n.foo as foo, count(*) AS count
WITH barCounts, collect({foo:foo, count:count}) as fooCounts
RETURN barCounts, fooCounts
Since you are trying to aggregate separate query results, you can also use UNION as a quick and easy way to return both at the same time.
Match (n)
RETURN "foo" as type, n.foo as value, count(*) AS count
UNION ALL
Match (n)
RETURN "bar" as type, n.bar as value, count(*) AS count
Just a few notes, both returns for a UNION must have the same column names.
Also, the "type" column in the example isn't necessary, but it shows how you can add filler if both queries don't have the same number of return columns. (Or if you want to tell which query the result is from.) If there is a "foo" and a "bar" with the same value+count, UNION ALL will keep both, and UNION will drop the duplicate (if you remove the type column).
Maybe it's outdated, but just in case someone needs it, I've found another approach using an APOC function which avoids running multiple times the same MATCH (n). In your case, it could be something like:
MATCH (n)
WITH collect(n.bar) as bars, collect(n.foo) as foos
WITH apoc.coll.frequenciesAsMap(bars) as barCounts, apoc.coll.frequenciesAsMap(foos) as fooCounts
RETURN barCounts, fooCounts
Single MATCH, multiple Counts.
Wish it could help someone!
Related
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 am sending a cypher query through php.
match (n:person)-[:watched]->(m:movie)
where m.Title in $mycollection
return count(distinct n.id);
this returns the number of people who have watched movies in my collection.
I actually want to return the list of names, and return n.name works fine.
When I try to return n.name and count(distinct n.id) at the same time, I lose the total count and get the count per row.
match (n:person)-[:watched]->(m:movie)
where m.Title in $mycollection
return n.name, count(distinct n.id);
does not work. The count column appears as 1 for each row.
As I'm using php, I've also tried:
$count = $result->getNodesCount();
to no avail. So I'm using php to count the array. But it feels like Cypher should be able to do it, right?
return n.name, count(distinct n.name) means "return each distinct n.name value and its number of distinct values". The number must always be 1, since a distinct value is, obviously, distinct.
If you are actually looking for the number of times each person had an outgoing relationship to a movie whose title is in $mycollection, do this instead (where count(*) counts the number of times a given n.name was matched):
MATCH (n:person)-->(m:movie)
WHERE m.Title in $mycollection
RETURN n.name, count(*);
Note that the above query omits the [watched] pattern found in your query, since that syntax (with no colon before watched) does no filtering at all. It merely assigns the relationship to a variable named watched, but that variable is not otherwise used, and is therefore superfluous.
If you had intended to use watched as the relationship type, then do this instead:
MATCH (n:person)-[:watched]->(m:movie)
WHERE m.Title in $mycollection
RETURN n.name, count(*);
This modified query returns the number of times each person watched a movie whose title is in $mycollection
I have the following records in my neo4j database
(:A)-[:B]->(:C)-[:D]->(:E)
(:C)-[:D]->(:E)
I want to get all the C Nodes and all the relations and related Nodes. If I do the query
Match (p:A)-[o:B]->(i:C)-[u:D]->(y:E)
Return p,o,i,u,y
I get the first to match if I do
Match (i:C)-[u:D]->(y:E)
Return i,u,y
I get the second to match.
But I want both of them in one query. How do I do that?
The easiest way is to UNION the queries, and pad unused variables with null (because all cyphers UNION'ed must have the same return columns
Match (p:A)-[o:B]->(i:C)-[u:D]->(y:E)
Return p,o,i,u,y
UNION
Match (i:C)-[u:D]->(y:E)
Return NULL as p, NULL as o,i,u,y
In your example though, the second match actually matches the last half of the first chain as well, so maybe you actually want something more direct like...
MATCH (c:C)
OPTIONAL MATCH (connected)
WHERE (c)-[*..20]-(connected)
RETURN c, COLLECT(connected) as connected
It looks like you're being a bit too specific in your query. If you just need, for all :C nodes, the connected nodes and relationships, then this should work:
MATCH (c:C)-[r]-(n)
RETURN c, r, n
I've got a graph where each node has label either A or B, and an index on the id property for each label:
CREATE INDEX ON :A(id);
CREATE INDEX ON :B(id);
In this graph, I want to find the node(s) with id "42", but I don't know a-priori the label. To do this I am executing the following query:
MATCH (n {id:"42"}) WHERE (n:A OR n:B) RETURN n;
But this query takes 6 seconds to complete. However, doing either of:
MATCH (n:A {id:"42"}) RETURN n;
MATCH (n:B {id:"42"}) RETURN n;
Takes only ~10ms.
Am I not formulating my query correctly? What is the right way to formulate it so that it takes advantage of the installed indices?
Here is one way to use both indices. result will be a collection of matching nodes.
OPTIONAL MATCH (a:B {id:"42"})
OPTIONAL MATCH (b:A {id:"42"})
RETURN
(CASE WHEN a IS NULL THEN [] ELSE [a] END) +
(CASE WHEN b IS NULL THEN [] ELSE [b] END)
AS result;
You should use PROFILE to verify that the execution plan for your neo4j environment uses the NodeIndexSeek operation for both OPTIONAL MATCH clauses. If not, you can use the USING INDEX clause to give a hint to Cypher.
You should use UNION to make sure that both indexes are used. In your question you almost had the answer.
MATCH (n:A {id:"42"}) RETURN n
UNION
MATCH (n:B {id:"42"}) RETURN n
;
This will work. To check your query use profile or explain before your query statement to check if the indexes are used .
Indexes are formed and and used via a node label and property, and to use them you need to form your query the same way. That means queries w/out a label will scan all nodes with the results you got.
Let's say we have the example query from the documentation:
MATCH (n:Actor)
RETURN n.name AS name
UNION
MATCH (n:Movie)
RETURN n.title AS name
I know that if I do that:
MATCH (n:Actor)
RETURN n.name AS name
LIMIT 5
UNION
MATCH (n:Movie)
RETURN n.title AS name
LIMIT 5
I can reduce the returned results of each sub query to 5.How can I LIMIT the total results of the union query?
This is not yet possible, but there is already an open neo4j issue that requests the ability to do post-UNION processing, which includes what you are asking about. You can add a comment to that neo4j issue if you support having it resolved.
This can be done using UNION post processing by rewriting the query using the COLLECT function and the UNWIND clause.
First we turn the columns of a result into a map (struct, hash, dictionary), to retain its structure. For each partial query we use the COLLECT to aggregate these maps into a list, which also reduces our row count (cardinality) to one (1) for the following MATCH. Combining the lists is a simple list concatenation with the “+” operator.
Once we have the complete list, we use UNWIND to transform it back into rows of maps. After this, we use the WITH clause to deconstruct the maps into columns again and perform operations like sorting, pagination, filtering or any other aggregation or operation.
The rewritten query will be as below:
MATCH (n:Actor)
with collect ({name: n.title}) as row
MATCH (n:Movie)
with row + collect({name: n.title}) as rows
unwind rows as row
with row.name as name
return name LIMIT 5
This is possible in 4.0.0
CALL {
MATCH (p:Person) RETURN p
UNION
MATCH (p:Person) RETURN p
}
RETURN p.name, p.age ORDER BY p.name
Read more about Post-union processing here https://neo4j.com/docs/cypher-manual/4.0/clauses/call-subquery/