NEO4J - 查找节点的所有入边和出边

Question

我希望找出所有出边和入边的数量之差，这些边连接的是城市节点，表示它们之间的转移关系。

我已经尝试了以下方法：

MATCH ()-[i:TRANSFERS]->(n:City {name:"London"}),(n:City {name:"London"})-[o:TRANSFERS]->() 
RETURN distinct n.name, count(i) AS incoming, count(o) as outgoing, count(o)-count(i) AS difference
ORDER BY outgoing - incoming DESC

并且还有这个：

MATCH ()-[i:TRANSFERS]->(n:City {name:"London"})
OPTIONAL MATCH (n:City {name:"London"})-[o:TRANSFERS]->() 
RETURN distinct n.name, count(i) AS incoming, count(o) as outgoing, count(o)-count(i) AS difference
ORDER BY outgoing - incoming DESC

但它们似乎没有起作用。有什么想法吗？

- SegFault

Answer 1

您可以在入站和出站连接的关系模式上使用大小：

MATCH (city:City {name:"London"})
WITH size((city)-[:TRANSFERS]->()) as out, 
     size((city)<-[:TRANSFERS]-()) as in,
     city
RETURN city, in, out, (out - in) as diff
ORDER BY diff DESC

- Christophe Willemsen

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它返回什么？ - Christophe Willemsen

出边和入边的数量与图中的不对应！如果我进行单独查询，数字会更大... - SegFault

你能否在浏览器中截取你的数据屏幕截图或提供一些数据？根据你的描述，这个查询肯定是有效的。 - Christophe Willemsen

也许一些关系名称不是TRANSFERS？ - Brian Underwood

还有，也许Neo4j版本有所不同？size是在2.3中引入的吗？ - Brian Underwood

我最喜欢的密码子语句 :) :) - Christophe Willemsen

Answer 2

我认为你需要分别匹配输入和输出（这里借用了Christophe的答案）：

MATCH (city:City {name:"London"})
WITH city, size((city)-[:TRANSFERS]->()) as out
WITH city, out, size((city)<-[:TRANSFERS]-()) as in,
RETURN city, in, out, (out - in) as diff
ORDER BY diff DESC

- Brian Underwood

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哈哈 :) 我发现了在 IN 谓词中的变量控制台转换，新的控制台地址 http://console.neo4j.org/r/ih4epf - Christophe Willemsen