关于neo4j图数据库笔记六-电影库和最短路径问题

知识图谱在工商企业、交往圈模型、系统架构、血缘关系、关联聚合场景、区域最短路径上都能发挥很大的作用，本笔记也只是简单的介绍了一下，介绍到此为止。

上述的场景的数据来源都来自于关系型数据，想让知识图谱发挥更大的作用，就必须靠自然语言处理，挖掘文本内部的关系，从而把非结构化数据转换为半结构化或结构化数据；目前还在学习中。

创建电影相关的演员、导演、制片商、作家和相关关系，这些数据来自于neo4j的movie数据

ACTED_IN（角色扮演）关系，共172条，源数据为演员，目标数据为电影，属性包括 roles，属性值为数组

DIRECTED（导演）关系，共44条，源数据为导演，目标数据为电影

PRODUCED（制片）关系，共15条，源数据为制片商，目标数据为电影

WROTE（写作）关系，共10条，源数据为作家，目标数据为电影

FOLLOWS（估计为回复）关系，共3条，源数据为影评人，目标数据为影评人

REVIEWED(影评)关系，共9条，源数据为影评人，目标数据为电影，属性包括 summary和rating

CREATE
  (Keanu)-[:ACTED_IN {roles:['Neo']}]->(TheMatrix),
  (Carrie)-[:ACTED_IN {roles:['Trinity']}]->(TheMatrix),
  (Laurence)-[:ACTED_IN {roles:['Morpheus']}]->(TheMatrix),
  (LanaW)-[:DIRECTED]->(TheMatrix),
  (JoelS)-[:PRODUCED]->(TheMatrix),
  (RobR)-[:DIRECTED]->(AFewGoodMen),
  (AaronS)-[:WROTE]->(AFewGoodMen)
CREATE
  (JamesThompson)-[:FOLLOWS]->(JessicaThompson),
  (PaulBlythe)-[:FOLLOWS]->(AngelaScope)
CREATE
  (JessicaThompson)-[:REVIEWED {summary:'An amazing journey', rating:95}]->(CloudAtlas),
  (JessicaThompson)-[:REVIEWED {summary:'You had me at Jerry', rating:92}]->(JerryMaguire)

1、查找所有相关电影、人物、关系

MATCH (n) RETURN n

2、查找名为“Tom Hanks”的人物

MATCH (person{name:"Tom Hanks"}) RETURN person

查找名为“Cloud Atlas”的电影

MATCH (movie{title:"Cloud Atlas"}) RETURN movie

3、随机查找10个人物的人名

MATCH (people:Person) RETURN people.name LIMIT 10

4、随机查找10个人物

MATCH (people:Person) RETURN people LIMIT 10

5、查找1990-2000年之间发行的电影名称

MATCH (nineties:Movie)

WHERE nineties.released>1990 AND nineties.released<2000

RETURN nineties.title,nineties.released

ORDER BY nineties.released

6、查找"Tom Hanks"参演过的电影名称

MATCH (people:Person{name:"Tom Hanks"})-[:ACTED_IN]->(actorMovies)

RETURN people,actorMovies

7、查找谁导演了电影“Cloud Atlas”

MATCH (move {title:"Cloud Atlas"}) <- [:DIRECTED] - (directors)

RETURN directors.name

8、查找与Tom Hanks一同出演过电影的人

MATCH (people:Person{name:"Tom Hanks"})-[:ACTED_IN]->(m)<-[:ACTED_IN]-(acActors)

RETURN acActors.name,m.title

查找与Tom Hanks一同出演过电影的人、电影

MATCH (people:Person{name:"Tom Hanks"})-[:ACTED_IN]->(m)<-[:ACTED_IN]-(acActors)

RETURN people,m,acActors

9、查找与电影"Cloud Atlas"相关的人

MATCH (people:Person) - [relatedTo] - (movie:Movie {title:"Cloud Atlas"})

RETURN people.name,Type(relatedTo),relatedTo,movie.title

10、查找与演员"Kevin Bacon"存在4条及以内关系的任何演员和电影

MATCH (people:Person{name:"Kevin Bacon"}) - [*1..4] - (hollywood)

RETURN DISTINCT hollywood

11、查找与演员"Kevin Bacon"与"Meg Ryan"之间的最短关系路径

MATCH p=shortestPath((people:Person{name:"Kevin Bacon"}) - [*] - (other:Person{name:"Meg Ryan"}))

RETURN p

12、查找没有与Tom Hanks合作过的演员

1、先找到Tom Hanks还没有合作过的演员，但Tom Hanks的合作伙伴曾经与其合作过

2、找到一个可以向他的潜在合作者介绍Tom Hanks的人

MATCH (people:Person{name:"Tom Hanks"}) - [:ACTED_IN]->(m)<-[:ACTED_IN]-(others),

(others)-[:ACTED_IN]->(m2)<-[:ACTED_IN]-(otherothers)

WHERE NOT (people)- [:ACTED_IN]->(m2)

RETURN otherothers.name AS Recommended,count(*) AS Strength

ORDER BY Strength DESC

13、找人将Tom Hanks介绍给Tom Cruise

MATCH

(tom:Person{name:"Tom Hanks"})-[:ACTED_IN]->(m)<-[:ACTED_IN]-(others),

(others)-[:ACTED_IN]->(m2)<-[:ACTED_IN]-(cruise:Person{name:"Tom Cruise"})

RETURN tom,m,others,m2,cruise

14、匹配某个电影相关的演员和导演等多种关系

MATCH (movie:Movie{title:'The Matrix Reloaded'})<-[:ACTED_IN|DIRECTED]-(person)

RETURN person.name

15、单条最短路径shortestPath函数

MATCH (apeople:Person{name:'Tom Hanks'}),(bpeople:Person{name:'Tom Cruise'}),

p=shortestPath((apeople)-[*..4]-(bpeople))

RETURN p

16、所有最短路径allshortestPaths函数

MATCH (apeople:Person{name:'Tom Hanks'}),(bpeople:Person{name:'Tom Cruise'}),

p=allshortestPaths((apeople)-[*..4]-(bpeople))

RETURN p

应用场景：根据关系属性寻找最短路径

CREATE 
(A:Node{name:"A"}),
(B:Node{name:"B"}),
(C:Node{name:"C"}),
(D:Node{name:"D"}),
(E:Node{name:"E"}),
(F:Node{name:"F"}),
(G:Node{name:"G"}),
(H:Node{name:"H"}),
(I:Node{name:"I"}),
(A)-[:path{dist:2}]->(B),
(A)-[:path{dist:3}]->(C),
(A)-[:path{dist:4}]->(D),
(B)-[:path{dist:3}]->(C),
(D)-[:path{dist:3}]->(C),
(C)-[:path{dist:3}]->(E),
(C)-[:path{dist:4}]->(F),
(C)-[:path{dist:11}]->(I),
(E)-[:path{dist:4}]->(F),
(E)-[:path{dist:4}]->(G),
(G)-[:path{dist:1}]->(H),
(H)-[:path{dist:2}]->(I)

找到A到I的最短路径，按照

1、单条最短路径shortestPath函数

MATCH (A:Node{name:'A'}),(I:Node{name:'I'}),

p=shortestPath((A)-[*..4]-(I))

RETURN p

2、所有最短路径allshortestPaths函数

MATCH (A:Node{name:'A'}),(I:Node{name:'I'}),

p=allshortestPaths((A)-[*..4]-(I))

RETURN p

3、根据距离计算最短路径

#根据dist值计算最短路径，得到最短路径的排序

MATCH (A:Node{name:'A'}),(I:Node{name:'I'}),p=((A)-[*1..8]->(I))

WITH REDUCE(tot=0, n IN rels(p)|tot + n.dist) AS ref, p

RETURN ref

ORDER BY ref ASC

#根据dist值计算最短路径，得到最短路径的图例，由于A-C-I有直接连通，所以这条线也出来了

MATCH (A:Node{name:'A'}),(I:Node{name:'I'}),p=((A)-[*1..8]->(I))

WITH REDUCE(tot=0, n IN rels(p)|tot + n.dist) AS ref, p

RETURN p,ref

ORDER BY ref ASC

LIMIT 1