[英]Neo4j cypher query from Training
我刚刚在http://www.neo4j.org/learn/online_course上完成了培训,并对实验室答案有几个问题。
首先是来自第2课的高级图形实验室。(未给出答案,并且未在图形小部件thingie中进行验证)
问题是:推荐基努·里维斯应与之合作的3位演员(但没有)。 提示是,您基本上应该选择与Keanu还没有ACTED_IN的电影具有ACTED_IN关系的三个人物。
该图包含具有ACTED_IN关系和DIRECTED关系的Person节点和Movie节点。
我想出了这个:
MATCH (a:Person)-[:ACTED_IN]->(movie:Movie)
WHERE NOT (:Person {name:"Keanu Reeves"})-[:ACTED_IN]->(movie)
RETURN a, count(movie)
ORDER BY count(movie) DESC
LIMIT 3
但我无法确定这是否真的排除了同一部电影,还是只排除了基努·里维斯(因为基恩努的电影中没有回返的演员,但无论如何他们可能都已经回了。
到目前为止,我已经找到了两个解决方案。
1:推荐最忙的演员基努·里夫斯(Keanu Reeves)未曾与之合作。
MATCH (p:Person)-[:ACTED_IN]->(m)
WHERE p.name <> 'Keanu Reeves'
AND NOT (p)-[:ACTED_IN]->()<-[:ACTED_IN]-(:Person{name:'Keanu Reeves'})
RETURN p.name, count(m) AS rating
ORDER BY count(m) DESC
LIMIT 3;
哪个产量
p.name | rating
--------------------------
Tom Hanks | 12
Meg Ryan | 5
Cuba Gooding Jr.| 4
2:推荐演员基努·里维斯(Keanu Reeves)的主演与大多数人合作
MATCH (f:Person)-[:ACTED_IN]->(m)<-[:ACTED_IN]-(c:Person),
(k:Person{name:'Keanu Reeves'})
WHERE c.name <> 'Keanu Reeves'
AND (f)-[:ACTED_IN]->()<-[:ACTED_IN]-(k)
AND NOT (c)-[:ACTED_IN]->()<-[:ACTED_IN]-(k)
RETURN c.name, count(c) AS Rating
ORDER BY Rating desc
LIMIT 3;
哪个产量
p.name | rating
--------------------------
Danny DeVito | 2
J.T. Walsh | 2
Tom Hanks | 2
今天我遇到了这个问题,在这里我做了什么
MATCH (keanu:Person)-[:ACTED_IN]->(movie),
(playedwith:Person)-[:ACTED_IN]->(movie),
(playedwith)-[t:ACTED_IN]->(othermovie),
(other:Person)-[:ACTED_IN]->(othermovie)
WHERE keanu.name = "Keanu Reeves"
AND NOT (other)-[:ACTED_IN]->(movie)
AND NOT (keanu)-[:ACTED_IN]->(othermovie)
RETURN other.name
,collect(DISTINCT othermovie)
,collect(DISTINCT playedwith)
,count(DISTINCT playedwith)
ORDER BY count(DISTINCT playedwith)desc
LIMIT 3
由于Distict太多,我虽然不喜欢它,但结果如下:
other.name | collect(DISTINCT othermovie) | collect(DISTINCT playedwith) | count(DISTINCT playedwith)
-----------------------------------------------------------------------------------------------------------------------------
Tom Hanks | ["Cloud Atlas", | ["Hugo Weaving","Charlize Theron"] | 2
| "That Thing You Do"] |
Tom Cruise | ["A Few Good Men"] | ["Jack Nicholson"] | 1
Robin Williams| ["The Birdcage"] | ["Gene Hackman"] | 1
因此,我发现了两种看起来不错的不同方法。 第一个查找具有“ ACTED_IN同一部电影”最多的人,其原始人不是与Keanu Reeves有“ ACTED_IN同一部电影”关系的人。
第二个查找的人尚未与Keanu Reeves交流过电影,但按照参与最多电影的人排序。
当然,在共享此关系的所有演员之间创建“ WORKED_WITH”关系,然后搜索Keanu尚未使用WORKED_WITH的所有人,这将是最简单的方法,但这使我猜错了。
第一个解决方案非常简单,而且看起来非常准确:
MATCH (a:Person {name:"Keanu Reeves"})-[:ACTED_IN]->(:Movie)<-[:ACTED_IN]-(b:Person)
WITH collect(b.name) AS FoF
MATCH (c:Person)-[:ACTED_IN]->(:Movie)<-[:ACTED_IN]-(d:Person)
WHERE not c.name IN FoF AND c.name <> "Keanu Reeves"
RETURN distinct c.name, count(distinct d)
ORDER BY count(distinct d) desc
limit 3
它返回:
c.name | count(distinct d)
-------------------------------
Tom Hanks | 34
Cuba Gooding Jr.| 24
Tom Cruise | 23
d是c拥有“ ACTED_IN”的人数。
编辑添加:
在给出答案后,我使用了更为简化的查询方法来提出以下建议:
MATCH (a:Person)-[:ACTED_IN]->()<-[:ACTED_IN]-(b:Person)
WHERE a.name <>'Keanu Reeves'
AND NOT (a)-[:ACTED_IN]->()<-[:ACTED_IN]-(b:Person {name:'Keanu Reeves'})
RETURN a.name, count(Distinct b) AS Rating
ORDER BY Rating DESC
LIMIT 3
返回与上面相同的内容。
或者,我将这些用于大多数电影中工作过的人:
MATCH (a:Person {name:"Keanu Reeves"})-[:ACTED_IN]->(:Movie)<-[:ACTED_IN]-(b:Person)
WITH collect(b.name) AS FoF
MATCH (c:Person)-[:ACTED_IN]->(m:Movie)<-[:ACTED_IN]-(d:Person)
WHERE not c.name IN FoF AND c.name <> "Keanu Reeves"
RETURN distinct c.name, count(distinct m)
ORDER BY count(distinct m) desc
limit 3
返回:
c.name | count(distinct m)
-------------------------------------------
Tom Hanks | 11
Meg Ryan | 5
Cuba Gooding Jr. | 4
其中m是他们制作的电影数量。
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