[英]Cypher diverging and converging path aggregate
我有一個問題,關於聚合在多個路徑上會聚然后又發散的問題。 然后,某些聚合應僅考慮路徑的一部分,而其他聚合則應考慮更多。
我將使用產品制造示例來盡力解釋這一點。 假設我有一家公司生產1種產品,其中某種材料由供應商提供。 更具體地說,該公司生產5種產品的一種產品,由10克材料組成。 因此,在制造過程中,他們使用了50克這種材料。 但是在生產過程中存在材料浪費,實際上他們消耗了70克,浪費了20克。
我要計算的是考慮到浪費的每種產品和供應商的物料校正重量。 在這種情況下,這很容易。 70克
現在,每個產品1和供應商1的物料1的更正重量為58.82克。 這是公式:
material composition = sum(production amount * product composition)
corrected weight = (production amount * product composition *
(purchased / (material composition)))
即
material composition = (5 * 10) + (20 * 40) = 850
corrected weight = (5 * 10 * (1000 / (850))) = 58.82
因此,在此示例上運行密碼查詢應該給我6個結果,因為這是產品,材料和供應商的排列數量。
問題是,如何編寫這樣的查詢。 我已經嘗試過減少函數,用-重復等,但是它似乎總是在錯誤的節點集上聚合。
為了完整起見,下面是生成圖的密碼:
創造:
create (c:Company {name:'test', id:'c1'}),
(p1:Product {name:'product1', id:'p1'}),
(p2:Product {name:'product2', id:'p2'}),
(m1:Material {name:'material1', id:'m1'}),
(m2:Material {name:'material2', id:'m2'}),
(s1:Supplier {name:'supplier1', id:'s1'}),
(s2:Supplier {name:'supplier2', id:'s2'}),
(s3:Supplier {name:'supplier3', id:'s3'})
角色:
match (c:Company {id:'c1'}),
(p1:Product {id:'p1'}),
(m1:Material {id:'m1'})
merge (c)<-[pb_r1:PRODUCED_BY {amount:5}]-(p1)-[co_r11:CONSISTS_OF {amount:10}]->(m1)
with c, p1, m1
match (p2:Product {id:'p2'})
merge (c)<-[pb_r2:PRODUCED_BY {amount:20}]-(p2)-[co_r12:CONSISTS_OF {amount:40}]->(m1)
with p1, p2, m1
match (s1:Supplier {id:'s1'})
merge (m1)-[pf_r1:PURCHASED_FROM {amount: 1000}]->(s1)
with p1, p2
match (m2:Material {id:'m2'})
merge (p1)-[co_r21:CONSISTS_OF {amount:30}]->(m2)
with p2, m2
merge (p2)-[co_r22:CONSISTS_OF {amount:80}]->(m2)
with m2
match (s2:Supplier {id:'s2'})
merge (m2)-[pf_r2:PURCHASED_FROM {amount: 1000}]->(s2)
with m2
match (s3:Supplier {id:'s3'})
merge (m2)-[pf_r3:PURCHASED_FROM {amount: 1000}]->(s3)
// Selection of the supply chain and production by Company
//
MATCH (C:Company {id:'c1'})
<-[pb:PRODUCED_BY]-
(P:Product)
-[co:CONSISTS_OF]->
(M:Material)
-[pf:PURCHASED_FROM]->
(S:Supplier)
// Grouping by materials, calculation material composition,
// and the preservation of the chain to the supplier
//
WITH M,
S, // group by supplier
SUM(pb.amount*co.amount) as mComp,
collect({
product:P,
prod: pb.amount,
comp: co.amount,
purchased: pf.amount
}) as tmps
// Calculating the correct weight by material and supplier
//
UNWIND tmps as tmp
RETURN M as material,
tmp['product'] as product,
S as supplier,
1.0 * tmp['prod'] * tmp['comp'] * tmp['purchased'] / mComp as cWeight
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.