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随着标签的引入，neo4j 获得了管理实体的简洁机制。现在，让我们假设我们有一个包含许多不同实体的通用图，并且想要找出哪些实体（标签）相互关联。这是一个起点： http ://console-test.neo4j.org/?id=wdnbuj

Graph Setup:
CYPHER 2.0
create 
(_1:Crew  {name:"Neo"}),
(_2:Crew  {name:"Morpheus"}),
(_3:Crew  {name:"Trinity"}),
(_4:Language  {name:"Cypher"}),
(_5:Machines  {name:"Agent Smith"}),
(_6:Machines  {name:"The Architect"}),
_1-[:KNOWS]->_2,
_1-[:LOVES]->_3,
_2-[:KNOWS]->_3,
_2-[:KNOWS]->_4,
_4-[:KNOWS]->_5,
_5-[:CODED_BY]->_6

Query:
CYPHER 2.0 
match x-[r]->y 
return head(labels(x)) as head, type(r), head(labels(y)) as tail

这将使我们对彼此相关的实体有一个概述：

+--------------------------------------+
| head       | type(r)    | tail       |
+--------------------------------------+
| "Machines" | "CODED_BY" | "Machines" |
| "Language" | "KNOWS"    | "Machines" |
| "Crew"     | "KNOWS"    | "Crew"     |
| "Crew"     | "KNOWS"    | "Language" |
| "Crew"     | "KNOWS"    | "Crew"     |
| "Crew"     | "LOVES"    | "Crew"     |
+--------------------------------------+

现在。是否有任何 Cypher 查询可以为这些三元组返回不同的值？

额外的问题：我们可以得到它们的计数/频率吗？

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1 回答 1

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我刚刚使用了您的起始查询，添加了 distinct 和 count。那是你要找的吗？或者您是否希望我从标签集合中拆分标签（如果没有某种展开功能，这并不容易）。

match x-[r]->y 
return distinct head(labels(x)) as head, type(r), head(labels(y)) as tail

match x-[r]->y 
return head(labels(x)) as head, type(r), head(labels(y)) as tail, count(*)
于 2013-05-17T11:33:57.787 回答