CODEWITHSUNDEEP
neo4jcypher
busyPixels
busyPixels

Reputation: 375

Get the count of multiple properties in neo4j

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.

Upvotes: 1

Views: 2593

Answers (3)

Sbunzini
Sbunzini

Reputation: 542

Maybe it's outdated, but just in case someone needs it, I've found another approach using an APOC function which avoids running multiple times the same MATCH (n).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!

Upvotes: 0

Tezra
Tezra

Reputation: 8833

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).

Upvotes: 3

InverseFalcon
InverseFalcon

Reputation: 30407

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

Upvotes: 4

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