Neo4j-需要聚合同一對節點之間不同關系上的值

Question

我必須將同一對節點之間的兩個不同關系和SUM（）的屬性相乘，然后根據值對對進行ORDER BY 。 例如：

Let (X:amps)-[:coocr{val:1}]->b
    (X:amps)-[:jacc{dist:2}]->b
    (X:amps)-[:coocr{val:2}]->c
    (X:amps)-[:jacc{dist:3}]->c
    (X:amps)-[:coocr]->d
    (X:amps)-[:jacc{dist:4}]->d
    (Y:amps)-[:coocr{val:2}]->b
    (Y:amps)-[:jacc{dist:3}]->b
    (Y:amps)-[:coocr{val:1}]->c
    (Y:amps)-[:jacc{dist:4}]->c
    (Y:amps)-[:coocr{val:4}]->d
    (Y:amps)-[:jacc{dist:3}]->d

現在，1）每個節點b，c，d與X，Y有兩個關系，它們分別相乘並成對相加2）返回最大值

我對問題的嘗試：

WITH [b,c,d] AS words                                                                
MATCH (i:amps) 
MATCH n where n.word in words
MATCH p=(i-[r:jaccard]->(n)) with i,r.dist as dist UNWIND dist as distances
MATCH q=(i-[s:coocr]->(n)) with distances,i,s.val as co UNWIND co as coocr
WITH i, SUM(distances*coocr) AS agg
RETURN i,agg ORDER BY agg DESC

我的匯總功能必須像這樣工作：

X-[]->b has[jaccard,coocr] , so, jacc.dist*coocr.val = 1*2 =2
X-[]->c has[jaccard,coocr] , so, jacc.dist*coocr.val = 2*3 =6
X-[]->d has[jaccard,] , so, jacc.dist*null = null*2 =0
                                                    --------------
                                                      sum  = 8
                                                    --------------

Y-[]->b has[jaccard,coocr] , so, jacc.dist*coocr.val = 4*2 =8
Y-[]->c has[jaccard,coocr] , so, jacc.dist*coocr.val = 1*3 =3
Y-[]->d has[jaccard,coocr] , so, jacc.dist*coocr.val = 4*4 =16
                                                    --------------
                                                      sum  = 27
                                                    --------------

然后按順序返回這些聚合。

我的查詢返回的聚合只是產品，而不是總和。 需要實現其總和。

Answer 1

那這個呢？

MATCH (a:amps)
OPTIONAL MATCH (a)-[coocr:coocr]-(n)
OPTIONAL MATCH (a)-[jacc:jacc]-(n)
WITH a, COALESCE(coocr.val, 0) AS val, COALESCE(jacc.dist, 0) AS dist
RETURN a, SUM(val * dist) AS agg
ORDER BY agg DESC

Answer 2

WITH ["best", "high", "quality","4k"] AS words                                                                
MATCH (i:amps) 
MATCH n where n.word in words
OPTIONAL MATCH p=(i-[r:jaccard]->(n)) with n,i,COALESCE(r.dist, 0)  as distances
OPTIONAL MATCH q=(i-[s:coocr]->(n)) with distances,n,i,COALESCE(s.val, 0) AS  coocr
WITH i,n,distances,coocr, (distances*coocr) AS agg
WITH i,SUM(agg) AS agg
RETURN i,agg

聚合功能，最后只用一行就解決了這個問題。 感謝您的建議Brain。