Neo4j-需要聚合同一对节点之间不同关系上的值

Question

我必须将同一对节点之间的两个不同关系和SUM（）的属性相乘，然后根据值对对进行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

现在，1）每个节点b，c，d与X，Y有两个关系，它们分别相乘并成对相加2）返回最大值

我对问题的尝试：

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

我的汇总功能必须像这样工作：

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
                                                    --------------

然后按顺序返回这些聚合。

我的查询返回的聚合只是产品，而不是总和。 需要实现其总和。

Answer 1

那这个呢？

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，最后只用一行就解决了这个问题。 感谢您的建议Brain。