使用子图聚合的递归查询

Question

我正在尝试使用Neo4j编写一个沿特定子图聚合数量的查询。

我们有两家商店Store1和Store2一家与供应商S1，另一家与供应商S2。我们将Store1的100个单位移至Store3，将Store2的200个单位移至Store3。

然后，我们将Store3的100个单位移至Store4。现在Store4有100个单位，大约33个来自供应商S1，66来自供应商S2。

我需要查询才能有效地返回此信息，例如

  

S1,33
  S2,66

我有一个递归查询来聚合每条路径上的所有移动

MATCH p=(store1:Store)-[m:MOVE_TO*]->(store2:Store { Name: 'Store4'}) 
RETURN store1.Supplier, reduce(amount = 0, n IN relationships(p) | amount + n.Quantity) AS reduction

返回：

| store1.Supplier |减少|
| -------------------- | ------------- |
| S1 | 200 |
| S2 | 300 |
| null | 100 |

所需

| store1.Supplier |减少|
| --------------------- | ------------- |
| S1 | 33.33 |
| S2 | 66.67 |

Answer 1

这个怎么样：

MATCH (s:Store) WHERE s.name = 'Store4'
MATCH (s)<-[t:MOVE_TO]-()<-[r:MOVE_TO]-(supp)
WITH t.qty as total, collect(r) as movements
WITH total, movements, reduce(totalSupplier = 0, r IN movements | totalSupplier + r.qty) as supCount
UNWIND movements as movement
RETURN startNode(movement).name as supplier, round(100.0*movement.qty/supCount) as pct

返回：

supplier    pct
Store1  33
Store2  67
Returned 2 rows in 151 ms

Answer 2

所以以下内容非常难看，但它适用于您已经给出的示例。

MATCH (s4:Store { Name:'Store4' })<-[r1:MOVE_TO]-(s3:Store)<-[r2:MOVE_TO*]-(s:Store) 
WITH s3, r1.Quantity as Factor, SUM(REDUCE(amount = 0, r IN r2 | amount + r.Quantity)) AS Total
MATCH (s3)<-[r1:MOVE_TO*]-(s:Store) 
WITH s.Supplier as Supplier, REDUCE(amount = 0, r IN r1 | amount + r.Quantity) AS Quantity, Factor, Total 
RETURN Supplier, Quantity, Total, toFloat(Quantity) / toFloat(Total) * Factor as Proportion

我确信它可以改进。