Question

我想在Neo4j图中测试Triangle Clustering。这是一个示例：

CREATE(a:Person { name: "a" })-[:FRIENDS]->(b:Person {name : "b"}),
(a)-[:WORKS_AT]->(p:Business {name : "Mcdonalds"}),
(b)-[:WORKS_AT]->(p),
(c:Person { name: "c"})-[:FRIENDS]->(a),
(c:Person { name: "c"})-[:FRIENDS]->(b),
(d:Person { name: "d"})-[:FRIENDS]->(a)
return *

MATCH (c:Person {name: "c"}),(p:Business {name : "Mcdonalds"}), (d:Person { name: "d"}),(b:Person {name : "b"})
CREATE (c)-[:WORKS_AT]->(p),
(e:Person { name: "e"})-[:FRIENDS]->(c),
(d)-[:FRIENDS]->(c),
(d)-[:FRIENDS]->(e),
(f:Person { name: "f"})-[:FRIENDS]->(b),
(g:Person { name: "g"})-[:FRIENDS]->(b),
(i:Person { name: "i"})-[:FRIENDS]->(b),
(h:Person { name: "h"})-[:FRIENDS]->(b),
(j:Person { name: "j"})-[:FRIENDS]->(b),
(k:Person { name: "k"})-[:FRIENDS]->(b)
return *

MATCH (g:Person {name: "g"}),(f:Person {name: "f"}),(c:Person {name: "c"}), (e:Person {name: "e"})
CREATE (g)-[:FRIENDS]->(c),
(f)-[:FRIENDS]->(c),
(g)-[:FRIENDS]->(e)
return *

在我的示例图中，我希望根据节点与麦当劳的：works_at关系选择节点a，b，c，然后查看那些具有：friends关系的节点并使用它们进行三角计数。我得到了部分答案：

CALL algo.triangleCount(
  'MATCH (p:Person)-[]-(:Person)-[:WORKS_AT]-(:Business {name : "Mcdonalds"}) RETURN id(p) as id',
  'MATCH (p1:Person)-[:FRIENDS]->(p2:Person) RETURN id(p1) as source, id(p2) as target',
  {concurrency:4, write:true, writeProperty:'triangle',graph:'cypher', clusteringCoefficientProperty:'coefficient'})
YIELD loadMillis, computeMillis, writeMillis, nodeCount, triangleCount, averageClusteringCoefficient

但是我想更接近documentation的流示例中列出的内容，并细分了nodeId（在此示例中为node.name），三角形和系数。

我已经接近：

CALL algo.triangleCount.stream(
  'MATCH (p:Person)-[]-(:Person)-[:WORKS_AT]-(:Business {name : "Mcdonalds"}) RETURN id(p) as id',
  'MATCH (p1:Person)-[:FRIENDS]->(p2:Person) RETURN id(p1) as source, id(p2) as target',
  {concurrency:4, write:true, writeProperty:'triangle',graph:'cypher', clusteringCoefficientProperty:'coefficient'})
YIELD nodeId, triangles, coefficient 
MATCH (p:Person) WHERE id(p) = nodeId
RETURN p.id as name, triangles, coefficient  ORDER BY coefficient DESC

Answer 1

CALL algo.triangleCount.stream('match (p:Person)-[*1..2]-(b:Business) return p', '[]', {concurrency:4})
YIELD nodeId, triangles, coefficient
MATCH (p:Person) WHERE id(p) = nodeId
RETURN p.name AS name, triangles, coefficient
ORDER BY triangles

这是我想出的答案。我所缺少的关键是了解triangleCount和triangleCount.stream之间的区别。流实际上是在分析数据，而普通的三角形计数仅提供性能，计数等统计信息。

三角形计数/聚类Neo4j

1 个答案: