Question

我必须在同一对节点之间将两个不同关系的属性相乘，然后 SUM（），然后 ORDER BY 基于该值的对。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具有与X，Y的两个关系，它们是相乘的并且成对地相加 2）返回最高值

我对这个问题的尝试：

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aggragate功能必须像：

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汇总。

我的查询返回聚合，这些聚合只是产品，而不是全部总和。需要达到其总和。

Answer 1

这个怎么样？

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

聚合函数，在最后但一行解决了问题。感谢Brain的建议。

Neo4j - 需要聚合同一对节点之间不同关系的值

2 个答案: