如何检查节点列表遵循neo4j中的有序路径？

1 回答

• 0

  首先，这里有一些看起来像你的样本的数据 . 我做了一些假设，列出的数据都是单独的链 .

  create (r1:Root {name: 'Root 1'})
  create (n1:Node {name: 'Neo4j'})
  create (n2:Node {name: 'Graph'})
  create (n3:Node {name: 'Database'})
  create r1-[:CONNECTED]->n1-[:CONNECTED]->n2-[:CONNECTED]->n3
  create (r2:Root {name: 'Root 2'})
  create (n4:Node {name: 'Neo4j'})
  create (n5:Node {name: 'Graph'})
  create (n6:Node {name: 'Search'})
  create r2-[:CONNECTED]->n4-[:CONNECTED]->n5-[:CONNECTED]->n6
  create (r3:Root {name: 'Root 3'})
  create (n7:Node {name: 'Neo4j'})
  create (n8:Node {name: 'Internals'})
  create r3-[:CONNECTED]->n7-[:CONNECTED]->n8
  return *
  

  现在，这是一个适用于有序列表的解决方案 . 0 ..使它能够使用只有一个节点的有序列表 .

  // start with the ordered list of node names
  with ['Neo4j', 'Graph', 'Database'] as ordered_list
  // start by matching paths that begin with the first node
  // in the ordered list and end with the last node
  match p=((a:Node {name: ordered_list[0]})-[:CONNECTED*0..]-(b:Node {name: ordered_list[length(ordered_list)-1]}))
  // collect these as the list of potential candidates
  with collect(p) as candidates, ordered_list
  // pass through the list and only keep the candidates that are 
  // the same length as the ordered list
  unwind candidates as candidate
  with case when (length(nodes(candidate)) = length(ordered_list)) 
    then candidate
  end as candidate, ordered_list
  // recollect the filtered candidates
  with collect(candidate) as candidates, ordered_list
  // pass through the candidates now again and compare the names
  // of the ordered list to the original ordered list
  // only keep those that match as matched
  unwind candidates as candidate
  with case 
    when reduce(names = [], n in nodes(candidate) | names + n.name) = ordered_list 
    then candidate
  end as matched
  // with the matched nodes find the root node that is attached to
  // first node in each matched path
  match (r:Root)--(head:Node)
  where id(head) = id(nodes(matched)[0])
  // return the root and the matched path nodes
  return r.name, nodes(matched)
  

  回复于 2024-04-29T22:57:12+08:00