Cypher查询如何按子节点类型稀有度排序优先返回稀有类型结果

MATCH (Parent)-[R]-(Child) WHERE ID(Parent)=$parentId
// 统计当前Parent下每个子节点标签的总数量，用于判定稀有度
WITH Parent, R, Child, count(*) OVER (PARTITION BY labels(Child)[0]) as tagTotalCount
// 保留原有子节点度数计算逻辑
CALL {
    WITH Child
    RETURN apoc.node.degree(Child) as ChildDegree
}
WITH Parent, Child, R, ChildDegree, tagTotalCount
// 优先按标签总数量升序排列，数量越少稀有度越高，同标签下保留原排序规则
RETURN Parent, Child, type(R), ChildDegree
ORDER BY tagTotalCount ASC, R
LIMIT 35

• 新增窗口函数count(*) OVER (PARTITION BY labels(Child)[0])统计当前Parent关联的所有子节点中，每个标签对应的总数量，存储为tagTotalCount字段
  • 若子节点存在多标签场景，可调整labels(Child)的取值逻辑，比如按指定标签优先级取值、或把全部标签拼接为分组key
• 排序逻辑调整为优先按tagTotalCount升序排列，标签总数量越小（稀有度越高）的子节点越靠前，保证稀有标签的结果优先被返回
• 同标签下的排序保留原有的R排序规则，也可根据业务需要调整为按ChildDegree等其他字段排序
• 最终返回结果会优先取完所有稀有标签的记录后，再取占比高的标签的记录，刚好满足示例中2条C、3条B、30条A的返回需求

MATCH (Parent)-[R]-(Child) WHERE ID(Parent)=$parentId
// 预计算每个标签的总数量与对应记录集合
WITH Parent, labels(Child)[0] as childTag, collect({child:Child, rel:R}) as tagRecords, count(*) as tagTotalCount
// 按标签数量升序排序，保证稀有标签在前
ORDER BY tagTotalCount ASC
// 展开所有记录，此时稀有标签的记录天然排在前面
UNWIND tagRecords as record
WITH Parent, record.child as Child, record.rel as R, tagTotalCount
CALL {
    WITH Child
    RETURN apoc.node.degree(Child) as ChildDegree
}
RETURN Parent, Child, type(R), ChildDegree
ORDER BY tagTotalCount ASC, R
LIMIT 35

内容的提问来源于stack exchange，提问作者ablaszkiewicz1