将2个节点之间的多个关系转换为具有权重的单个关系 [英] Convert multiple relationships between 2 nodes to a single one with weight

问题描述

我有以下图表，描述了在文档中同时出现的汽车品牌:

I have the following graph, describing co-occurrence of car brands in documents:

CREATE 
  (`0` :Car {value:"Ford"})
, (`1` :Car {value:"Subaru"})
, (`2` :Car {value:"VW"})
, (`0`)-[:`DOCUMENT` {value:"DOC-1"}]->(`1`)
, (`0`)-[:`DOCUMENT` {value:"DOC-2"}]->(`1`)
, (`1`)-[:`DOCUMENT` {value:"DOC-3"}]->(`2`);

如果两个节点之间有很多关系(出于可视化的目的)，我想将其替换为一个并计算权重:

If there are many relationships between two nodes - for the purpose of visualization - I want to replace it with a single one and calculate the weight:

VW ---1--- Subaru ---2--- Ford

如何实现?

我尝试了以下查询:

MATCH (n1)-[r1:DOCUMENT]-(n2)
RETURN n1, n2, apoc.create.vRelationship(n1, 'WEIGHT', {weight:count(r1)}, n2);

但这不是不是预期结果:

推荐答案

如果未在MATCH (n1)-[r1:DOCUMENT]-(n2) RETURN *;中指定关系方向，则返回每个关系两次:

If the relationship direction is not specified in MATCH (n1)-[r1:DOCUMENT]-(n2) RETURN *;, every relationship is returned twice:

╒══════════════════╤══════════════════╤═════════════════╕
│"n1"              │"n2"              │"r1"             │
╞══════════════════╪══════════════════╪═════════════════╡
│{"value":"Ford"}  │{"value":"Subaru"}│{"value":"DOC-1"}│
├──────────────────┼──────────────────┼─────────────────┤
│{"value":"Ford"}  │{"value":"Subaru"}│{"value":"DOC-2"}│
├──────────────────┼──────────────────┼─────────────────┤
│{"value":"Subaru"}│{"value":"Ford"}  │{"value":"DOC-1"}│
├──────────────────┼──────────────────┼─────────────────┤
│{"value":"Subaru"}│{"value":"Ford"}  │{"value":"DOC-2"}│
├──────────────────┼──────────────────┼─────────────────┤
│{"value":"Subaru"}│{"value":"VW"}    │{"value":"DOC-3"}│
├──────────────────┼──────────────────┼─────────────────┤
│{"value":"VW"}    │{"value":"Subaru"}│{"value":"DOC-3"}│
└──────────────────┴──────────────────┴─────────────────┘

通过将MATCH子句更改为MATCH (n1)-[r1:DOCUMENT]->(n2)或MATCH (n1)<-[r1:DOCUMENT]-(n2)来指定方向.查询的其余部分正确.

Specify the direction by changing the MATCH clause to either MATCH (n1)-[r1:DOCUMENT]->(n2) or MATCH (n1)<-[r1:DOCUMENT]-(n2). The remaining part of the query is correct.

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