图重构-重复关系重构

MERGE (A:Test {name:'A'})
MERGE (B:Test {name:'B'})
CREATE (A)-[:Realation {name:'属性1'}]->(B)
CREATE (A)-[:Realation {name:'属性1'}]->(B)
CREATE (A)-[:Realation {name:'属性2'}]->(B)
CREATE (A)-[:Realation {name:'属性2'}]->(B)
CREATE (A)-[:Realation {name:'属性2'}]->(B)
CREATE (A)-[:Realation {name:'属性3'}]->(B)

MATCH p=(A:Test {name:'A'})-->(B:Test {name:'B'}) RETURN p

想请教下大佬，如何删除两个节点间的重复关系，重复的定义指的是，关系的属性不同
比如
(A)-[:Realation{name:‘属性1’]-(B)
(A)-[:Realation{name:‘属性1’]-(B)
(A)-[:Realation{name:‘属性2’]-(B)
(A)-[:Realation{name:‘属性2’]-(B)
(A)-[:Realation{name:‘属性2’]-(B)
(A)-[:Realation{name:‘属性3’]-(B)
想把重复的部分去掉，就是变成
(A)-[:Realation{name:‘属性1’]-(B)
(A)-[:Realation{name:‘属性2’]-(B)
(A)-[:Realation{name:‘属性3’]-(B)

MATCH p=(A:Test {name:'A'})-[r]->(B:Test {name:'B'})
WITH ID(r) AS id,r.name AS name
WITH name,COLLECT(id) AS relIds
WITH name,relIds,SIZE(relIds) AS relIdsSize
WHERE relIdsSize>1
WITH name,apoc.coll.subtract(relIds, [relIds[0]]) AS deleteRelIds
WITH name,deleteRelIds
MATCH ()-[r]-() WHERE ID(r) IN deleteRelIds DELETE r

CALL apoc.do.case([relationship=1,\'MATCH (from:Label {hcode:$fromHcode}),(to:Label {hcode:$toHcode}) MERGE (from)-[:NEXT]->(to)\',relationship=-1,\'MATCH (from:Label {hcode:$fromHcode}),(to:Label {hcode:$toHcode}) MERGE (from)<-[:NEXT]-(to)\'],\'\',{fromHcode:fromHcode,toHcode:toHcode}) YIELD value RETURN value

http://neo4j.com.cn/topic/5f3b28e4a4477ec754d2b55f

https://www.lyonwj.com/