neo4j指所有朋友的财产

Question

有一个图形，如：

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CREATE (Alice:Person {id:'a', fraud:1})
    CREATE (Bob:Person {id:'b', fraud:0})
    CREATE (Charlie:Person {id:'c', fraud:0})
    CREATE (David:Person {id:'d', fraud:0})
    CREATE (Esther:Person {id:'e', fraud:0})
    CREATE (Fanny:Person {id:'f', fraud:0})
    CREATE (Gabby:Person {id:'g', fraud:0})
    CREATE (Fraudster:Person {id:'h', fraud:1})


CREATE
  (Alice)-[:CALL]->(Bob),
  (Bob)-[:SMS]->(Charlie),
  (Charlie)-[:SMS]->(Bob),
  (Fanny)-[:SMS]->(Charlie),
  (Esther)-[:SMS]->(Fanny),
  (Esther)-[:CALL]->(David),
  (David)-[:CALL]->(Alice),
  (David)-[:SMS]->(Esther),
  (Alice)-[:CALL]->(Esther),
  (Alice)-[:CALL]->(Fanny),
  (Fanny)-[:CALL]->(Fraudster)

当尝试查询时，例如：

MATCH (a)-->(b)
WHERE b.fraud = 1
RETURN (count() / ( MATCH (a) -->(b) RETURN count() ) * 100)

我想计算一个用户的欺诈性，该用户(欺诈只定义为0或1被定义为所有连接节点欺诈级别的平均值)：

MATCH ()--(f)
RETURN f.id, f.fraud, COUNT(*), COLLECT(f) AS fs

返回正确数量的朋友，但无法访问这些，即在收集语句中只访问节点本身：

╒══════╤═════════╤══════════════╤══════════╤══════════════════════════════════════════════════════════════════════╕
│"f.id"│"f.fraud"│"avg(f.fraud)"│"COUNT(*)"│"fs"                                                                  │
╞══════╪═════════╪══════════════╪══════════╪══════════════════════════════════════════════════════════════════════╡
│"h"   │1        │1             │1         │[{"fraud":1,"id":"h"}]                                                │
├──────┼─────────┼──────────────┼──────────┼──────────────────────────────────────────────────────────────────────┤
│"f"   │0        │0             │4         │[{"fraud":0,"id":"f"},{"fraud":0,"id":"f"},{"fraud":0,"id":"f"},{"frau│
│      │         │              │          │d":0,"id":"f"}]                                                       │
....

也就是天真地计算平均值

MATCH ()--(f)
RETURN f.id, avg(f.fraud)

只考虑这个节点，而不考虑网络。如何考虑节点的社交网络而不是(达到定义的深度，即这里的1)来改进neo4j percentage of attribute for social network的原始答案

编辑

MATCH p = ()--()
UNWIND nodes(p) AS f
RETURN f.id, f.fraud, COUNT(*), COLLECT({id: f.id, fraud: f.fraud}) AS fs

将只返回列表中原始节点的副本，而不返回连接的节点：

│"f.id"│"f.fraud"│"COUNT(*)"│"fs"                                                                  │
╞══════╪═════════╪══════════╪══════════════════════════════════════════════════════════════════════╡
│"h"   │1        │2         │[{"id":"h","fraud":1},{"id":"h","fraud":1}]                           │
├──────┼─────────┼──────────┼──────────────────────────────────────────────────────────────────────┤
│"f"   │0        │8         │[{"id":"f","fraud":0},{"id":"f","fraud":0},{"id":"f","fraud":0},{"id":│
│      │         │          │"f","fraud":0},{"id":"f","fraud":0},{"id":"f","fraud":0},{"id":"f","fr│
│      │         │          │aud":0},{"id":"f","fraud":0}]                                         │

编辑2

MATCH p = (source)--(destination)
RETURN source.id, source.fraud, COUNT(*), COLLECT({id: destination.id, fraud: destination.fraud}) AS neighbors

已经很接近了，但是缺少avg功能

Answer 1

MATCH p = (source)-[*..3]-(destination)
RETURN source.id, source.fraud, COUNT(*), avg(destination.fraud), COLLECT({id: destination.id, fraud: destination.fraud}) AS neighbors

包括定义为平均数的欺诈行为。