Question

我有一些数据，其中一些关系比其他关系重要，并加权来表明它。例如，我有这样的图：

  (city_a:City)-[:has{weight:10}]->(casino:Event)
  (city_a:City)-[:has{weight:1}]->(restaurant:Event)
  (city_a:City)-[:has{weight:30}]->(university:Event)    
  (city_a:City)-[:has{weight:25}]->(library:Event) 
   ......
  (city_b:City)-[:has{weight:2}]->(casino:Event)
  (city_b:City)-[:has{weight:2}]->(restaurant:Event)
  (city_b:City)-[:has{weight:5}]->(university:Event)    
  (city_b:City)-[:has{weight:10}]->(library:Event) 
   ......

我输入如下

  Input: {casino, restaurant, university}

我应该输出

  Output: city_a

作为答案，因为它的关系比其他城市的关系更多。

（也许我的图表模型不是正确的，但我也想不到什么。建议非常受欢迎）。

那么，如何为这种情况编写密码查询？

提前致谢！

Answer 1

以下是基于您的描述的示例图表，其中引入了“city_c”以涵盖所需输入事件少于一个城市的情况：

merge (city_a:City{name:'city_a'})
merge (city_b:City{name:'city_b'})
merge (city_c:City{name:'city_c'})

merge (casino:Event{name:'casino'})
merge (restaurant:Event{name:'restaurant'})
merge (university:Event{name:'university'})
merge (library:Event{name:'library'})

merge (city_a)-[:has{weight:10}]->(casino)
merge (city_a)-[:has{weight:1}]->(restaurant)
merge (city_a)-[:has{weight:30}]->(university)    
merge (city_a)-[:has{weight:25}]->(library) 

merge (city_b)-[:has{weight:2}]->(casino)
merge (city_b)-[:has{weight:2}]->(restaurant)
merge (city_b)-[:has{weight:5}]->(university)    
merge (city_b)-[:has{weight:10}]->(library) 

merge (city_c)-[:has{weight:100}]->(university)

我们还需要一个索引：事件（名称），用于通过输入字符串进行快速查找（您可能还需要索引：城市（名称），尽管这不需要或与此一起使用特别查询）

create index on :Event(name)

由此我们可以创建查询，以便根据输入事件与具有这些事件的城市进行匹配，并按每个城市的匹配事件数量和评级总和进行排序。

而不是with ... as input您想要参数化您的查询。

with ['casino', 'restaurant', 'university'] as input
match (e:Event)<-[r:has]-(city:City)
where e.name in input
with city, count(r) as eventCount, sum(r.weight) as weightSum
return city, eventCount, weightSum
order by eventCount desc, weightSum desc
limit 1

Neo4j：加权关系，如何反映查询中的重要关系？

1 个答案: