CODEWITHSUNDEEP
neo4jcypher
mixiul__
mixiul__

Reputation: 395

Find most common property - cypher

Is there a way I can use cypher to derive the most common (and second most common) property value of a particular node type?

At the moment I am identifying the different values for the property: 

MATCH(a:Label1)-[r:REL]->(b:Label2) RETURN DISTINCT b.prop;

And then counting them individually: 

MATCH(a:Label1)-[r:REL]->(b:Label2) WHERE b.prop = "x" RETURN COUNT(x);

Any help would be appreciated, I am using Neo4j 3.0.0.

I'm not sure how to combine these queries.

Upvotes: 4

Views: 1178

Answers (3)

Dave Bennett
Dave Bennett

Reputation: 11216

You can use the keys() function to get the properties from each matching node. This returns a collection of keys for that node. Unwind the collections and aggregate the counts for each property accross all matching nodes.

MATCH (a:Label1)-[r:REL]->(b:Label2)
UNWIND keys(b) as prop
RETURN prop, count(*) as num_props
ORDER BY num_props desc

Upvotes: 3

Christophe Willemsen
Christophe Willemsen

Reputation: 20185

It is as easy as :

MATCH (a:Label1)-[r:REL]->(b:Label2)
RETURN b.prop, count(*) as occurences
ORDER BY occurences DESC
LIMIT 2

You can read about automatic aggregation here : http://neo4j.com/docs/stable/query-aggregation.html

Upvotes: 7

stdob--
stdob--

Reputation: 29172

//
// Get patterns
MATCH (a:Label1)-[r:REL]->(b:Label2)
//
// Pass through all keys
UNWIND KEYS(b) as key
   //
   // Get patterns again
   MATCH (a:Label1)-[r:REL]->(b:Label2)
//
// Aggregating by title of the properties and its content
RETURN key as propName, b[key] as propValue, count(b[key]) as propRank
//
// Sort the largest property-value pairs first
ORDER BY propRank DESC

Upvotes: 2

Related Questions