具有子图聚合的递归查询(任意深度)

Question

我问一个问题前面关于沿着图形聚集量.提供的两个答案运作良好,但现在我正在尝试将Cypher查询扩展到可变深度的图表.

总结一下,我们从一堆叶子商店开始,这些叶子商店都与特定供应商相关联,这是Store节点上的一个属性.然后将库存移至其他商店,每个供应商的比例对应于他们对原始商店的贡献.

所以对于节点B02,S2贡献750/1250 = 60%和S3贡献40%.然后,我们600台我们对B02其中60%属于S2与40%对S3等.

我们想知道最终700个单位的百分比D01属于每个供应商.供应商名称相同的供应商.因此,对于上图,我们期望:

S1,38.09
S2,27.61
S3,34.28

我使用这个Cypher脚本编写了一个图表:

CREATE (A01:Store {Name: 'A01', Supplier: 'S1'})
CREATE (A02:Store {Name: 'A02', Supplier: 'S1'})
CREATE (A03:Store {Name: 'A03', Supplier: 'S2'})
CREATE (A04:Store {Name: 'A04', Supplier: 'S3'})
CREATE (A05:Store {Name: 'A05', Supplier: 'S1'})
CREATE (A06:Store {Name: 'A06', Supplier: 'S1'})
CREATE (A07:Store {Name: 'A07', Supplier: 'S2'})
CREATE (A08:Store {Name: 'A08', Supplier: 'S3'})

CREATE (B01:Store {Name: 'B01'})
CREATE (B02:Store {Name: 'B02'})
CREATE (B03:Store {Name: 'B03'})
CREATE (B04:Store {Name: 'B04'})

CREATE (C01:Store {Name: 'C01'})
CREATE (C02:Store {Name: 'C02'})

CREATE (D01:Store {Name: 'D01'})

CREATE (A01)-[:MOVE_TO {Quantity: 750}]->(B01)
CREATE (A02)-[:MOVE_TO {Quantity: 500}]->(B01)
CREATE (A03)-[:MOVE_TO {Quantity: 750}]->(B02)
CREATE (A04)-[:MOVE_TO {Quantity: 500}]->(B02)
CREATE (A05)-[:MOVE_TO {Quantity: 100}]->(B03)
CREATE (A06)-[:MOVE_TO {Quantity: 200}]->(B03)
CREATE (A07)-[:MOVE_TO {Quantity: 50}]->(B04)
CREATE (A08)-[:MOVE_TO {Quantity: 450}]->(B04)

CREATE (B01)-[:MOVE_TO {Quantity: 400}]->(C01)
CREATE (B02)-[:MOVE_TO {Quantity: 600}]->(C01)
CREATE (B03)-[:MOVE_TO {Quantity: 100}]->(C02)
CREATE (B04)-[:MOVE_TO {Quantity: 200}]->(C02)

CREATE (C01)-[:MOVE_TO {Quantity: 500}]->(D01)
CREATE (C02)-[:MOVE_TO {Quantity: 200}]->(D01)

目前的查询是这样的:

MATCH (s:Store { Name:'D01' })
MATCH (s)<-[t:MOVE_TO]-()<-[r:MOVE_TO]-(supp)
WITH t.Quantity as total, collect(r) as movements
WITH total, movements, reduce(totalSupplier = 0, r IN movements | totalSupplier + r.Quantity) as supCount
UNWIND movements as movement
RETURN startNode(movement).Supplier as Supplier, round(100.0*movement.Quantity/supCount) as pct

我正在尝试使用递归关系,类似于以下内容:

MATCH (s)<-[t:MOVE_TO]-()<-[r:MOVE_TO*]-(supp)

但是,它提供了到终端节点的多条路径,我需要在每个节点聚合库存.

Answer 1

此查询为符合问题中描述的模型的任意图形生成正确的结果。（当Storex 将商品移动到Storey 时，假设Supplier移动商品的百分比与 x 相同Store。）

然而，该解决方案不仅仅包含单个 Cypher 查询（因为这可能是不可能的）。相反，它涉及多个查询，其中一个查询必须迭代，直到计算级联通过整个节点图Store。该迭代查询将清楚地告诉您何时停止迭代。需要其他 Cypher 查询来：准备迭代图、报告“最终”节点的供应商百分比以及清理图（以便将其恢复到下面步骤 1 之前的状态） 。

这些查询可能可以进一步优化。

以下是所需的步骤：

1. 为迭代查询准备图形（初始化pcts所有起始Store节点的临时数组）。这包括创建一个单例Suppliers节点，该节点具有一个包含所有供应商名称的数组。这用于建立临时数组元素的顺序pcts，并将这些元素映射回正确的供应商名称。

   MATCH (store:Store)
   WHERE HAS (store.Supplier)
   WITH COLLECT(store) AS stores, COLLECT(DISTINCT store.Supplier) AS csup
   CREATE (sups:Suppliers { names: csup })
   WITH stores, sups
   UNWIND stores AS store
   SET store.pcts =
     EXTRACT(i IN RANGE(0,LENGTH(sups.names)-1,1) |
       CASE WHEN store.Supplier = sups.names[i] THEN 1.0 ELSE 0.0 END)
   RETURN store.Name, store.Supplier, store.pcts;
   

   这是问题数据的结果：

   +---------------------------------------------+
   | store.Name | store.Supplier | store.pcts    |
   +---------------------------------------------+
   | "A01"      | "S1"           | [1.0,0.0,0.0] |
   | "A02"      | "S1"           | [1.0,0.0,0.0] |
   | "A03"      | "S2"           | [0.0,1.0,0.0] |
   | "A04"      | "S3"           | [0.0,0.0,1.0] |
   | "A05"      | "S1"           | [1.0,0.0,0.0] |
   | "A06"      | "S1"           | [1.0,0.0,0.0] |
   | "A07"      | "S2"           | [0.0,1.0,0.0] |
   | "A08"      | "S3"           | [0.0,0.0,1.0] |
   +---------------------------------------------+
   8 rows
   83 ms
   Nodes created: 1
   Properties set: 9
   

2. 迭代查询（重复运行直到返回 0 行）

   MATCH p=(s1:Store)-[m:MOVE_TO]->(s2:Store)
   WHERE HAS(s1.pcts) AND NOT HAS(s2.pcts)
   SET s2.pcts = EXTRACT(i IN RANGE(1,LENGTH(s1.pcts),1) | 0)
   WITH s2, COLLECT(p) AS ps
   WITH s2, ps, REDUCE(s=0, p IN ps | s + HEAD(RELATIONSHIPS(p)).Quantity) AS total
   FOREACH(p IN ps |
     SET HEAD(RELATIONSHIPS(p)).pcts = EXTRACT(parentPct IN HEAD(NODES(p)).pcts | parentPct * HEAD(RELATIONSHIPS(p)).Quantity / total)
   )
   FOREACH(p IN ps |
     SET s2.pcts = EXTRACT(i IN RANGE(0,LENGTH(s2.pcts)-1,1) | s2.pcts[i] + HEAD(RELATIONSHIPS(p)).pcts[i])
   )
   RETURN s2.Name, s2.pcts, total, EXTRACT(p IN ps | HEAD(RELATIONSHIPS(p)).pcts) AS rel_pcts;
   

   迭代 1 结果：

   +-----------------------------------------------------------------------------------------------+
   | s2.Name | s2.pcts       | total | rel_pcts                                                    |
   +-----------------------------------------------------------------------------------------------+
   | "B04"   | [0.0,0.1,0.9] | 500   | [[0.0,0.1,0.0],[0.0,0.0,0.9]]                               |
   | "B01"   | [1.0,0.0,0.0] | 1250  | [[0.6,0.0,0.0],[0.4,0.0,0.0]]                               |
   | "B03"   | [1.0,0.0,0.0] | 300   | [[0.3333333333333333,0.0,0.0],[0.6666666666666666,0.0,0.0]] |
   | "B02"   | [0.0,0.6,0.4] | 1250  | [[0.0,0.6,0.0],[0.0,0.0,0.4]]                               |
   +-----------------------------------------------------------------------------------------------+
   4 rows
   288 ms
   Properties set: 24
   

   迭代 2 结果：

   +-------------------------------------------------------------------------------------------------------------------------------+
   | s2.Name | s2.pcts                                      | total | rel_pcts                                                     |
   +-------------------------------------------------------------------------------------------------------------------------------+
   | "C02"   | [0.3333333333333333,0.06666666666666667,0.6] | 300   | [[0.3333333333333333,0.0,0.0],[0.0,0.06666666666666667,0.6]] |
   | "C01"   | [0.4,0.36,0.24]                              | 1000  | [[0.4,0.0,0.0],[0.0,0.36,0.24]]                              |
   +-------------------------------------------------------------------------------------------------------------------------------+
   2 rows
   193 ms
   Properties set: 12
   

   迭代 3 结果：

   +---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
   | s2.Name | s2.pcts                                                       | total | rel_pcts                                                                                                                    |
   +---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
   | "D01"   | [0.38095238095238093,0.27619047619047615,0.34285714285714286] | 700   | [[0.2857142857142857,0.2571428571428571,0.17142857142857143],[0.09523809523809522,0.01904761904761905,0.17142857142857143]] |
   +---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+
   1 row
   40 ms
   Properties set: 6
   

   迭代 4 结果：

   +--------------------------------------+
   | s2.Name | s2.pcts | total | rel_pcts |
   +--------------------------------------+
   +--------------------------------------+
   0 rows
   69 ms
   

3. 列出结束节点Supplier的非零百分比。Store

   MATCH (store:Store), (sups:Suppliers)
   WHERE NOT (store:Store)-[:MOVE_TO]->(:Store) AND HAS(store.pcts)
   RETURN store.Name, [i IN RANGE(0,LENGTH(sups.names)-1,1) WHERE store.pcts[i] > 0 | {supplier: sups.names[i], pct: store.pcts[i] * 100}] AS pcts;
   

   结果：

   +----------------------------------------------------------------------------------------------------------------------------------+
   | store.Name | pcts                                                                                                                |
   +----------------------------------------------------------------------------------------------------------------------------------+
   | "D01"      | [{supplier=S1, pct=38.095238095238095},{supplier=S2, pct=27.619047619047617},{supplier=S3, pct=34.285714285714285}] |
   +----------------------------------------------------------------------------------------------------------------------------------+
   1 row
   293 ms
   

4. 清理（​​删除所有临时pcts道具和Suppliers节点）。

   MATCH (s:Store), (sups:Suppliers)
   OPTIONAL MATCH (s)-[m:MOVE_TO]-()
   REMOVE m.pcts, s.pcts
   DELETE sups;
   

   结果：

   0 rows
   203 ms
   +-------------------+
   | No data returned. |
   +-------------------+
   Properties set: 29
   Nodes deleted: 1