你需要为每个地点、设备和用户使用apoc.Coll.Companies获取一种物品组合。然后展开以组合每一项。结果是3x3+3x3+3x3+3x3x3=54行

MATCH (l:Location)
MATCH (d:Device)
MATCH (u:User)
WITH  collect(l) as locations, collect(d) as devices, collect(u) as users
WITH apoc.coll.combinations(locations,1,1) as locations,  apoc.coll.combinations(devices,1,1) as devices, apoc.coll.combinations(users,1,1) as users 
UNWIND locations as location
UNWIND devices as device
UNWIND users as user
CALL {
       WITH location, device 
       RETURN collect(distinct  location+device) as cols 

       UNION ALL

       WITH location, user
       RETURN collect(distinct  location+user) as cols 

       UNION ALL

       WITH device, user
       RETURN collect(distinct  device+user) as cols

       UNION ALL

       WITH location, device, user
       RETURN collect(distinct  location+device+user) as cols
}
RETURN collect(distinct cols) as combColl

当我展开最终集合时的结果：

╒══════════════════════════════╕
│"combColl"                    │
╞══════════════════════════════╡
│[[{"ID":1},{"ID":1}]]         │
├──────────────────────────────┤
│[[{"ID":1},{"ID":2}]]         │
├──────────────────────────────┤
│[[{"ID":1},{"ID":3}]]         │

....

├──────────────────────────────┤
│[[{"ID":3},{"ID":2},{"ID":3}]]│
├──────────────────────────────┤
│[[{"ID":3},{"ID":3},{"ID":1}]]│
├──────────────────────────────┤
│[[{"ID":3},{"ID":3},{"ID":2}]]│
├──────────────────────────────┤
│[[{"ID":3},{"ID":3},{"ID":3}]]│
└──────────────────────────────┘