Question

我有一个数据库，其中的节点和关系是这样创建的：

create ({foo: 1})-[:r]->({foo: 1}), ({foo: 1})-[:r]->({foo: 2}), ({foo: 1})-[:r]->({foo: 2})

对于每个路径，我可以通过以下方式生成属性列表：

MATCH path = (start)-[*]->(end)
WHERE NOT ()-->(start) and NOT (end)-->()
RETURN [node in nodes(path) | node.foo] as foos

返回

foos
------
[1, 1]
[1, 2]
[1, 2]

但是我只希望每个列表中都有不同的属性。

受到Cypher: Extract unique values from collection的启发

MATCH path = (start)-[*]->(end)
WHERE NOT ()-->(start) and NOT (end)-->()
WITH [node in nodes(path) | node.foo] as foos
UNWIND foos as x
RETURN COLLECT(DISTINCT x) as foos_unique

但这为我提供了所有节点的独特属性：

foos_unique
----
[1, 2]

如何从每个列表中提取不同的元素，这样结果将是

foos_unique
------
[1]
[1, 2]
[1, 2]

Answer 1

[已更新]

REDUCE函数将帮助：

MATCH path = (start)-[*]->(end)
WHERE NOT ()-->(start) and NOT (end)-->()
RETURN REDUCE(s = [], n in NODES(path) |
  CASE WHEN NOT EXISTS(n.foo) OR n.foo IN s THEN s ELSE s + n.foo END) as foos

Answer 2

这可以完成工作。无需收集唯一元素，我们可以像这样比较foo和start的{{1}}：

end

输出为：

MATCH (start)-[*]->(end)
WHERE NOT ()-->(start) and NOT (end)-->()
WITH start.foo as foos, end.foo as fooe 
RETURN case when foos = fooe then [foos] else [foos, fooe] end as foos_unique

在neo4j中，从列表中提取唯一元素

2 个答案: