密码找到关系方向

Question

如何找到与包含路径相关的关系方向？我需要这个来进行考虑关系方向的加权图搜索（用0 衡量“错误”方向，另请参见 cmets）。

让我们说：

START a=node({param})
MATCH a-[*]-b
WITH a, b
MATCH p = allshortestpaths(a-[*]-b)
RETURN extract(r in rels(p): flows_with_path(r)) as in_flow

在哪里 flows_with_path = 1 如果sp = (a)-[*0..]-[r]->[*0..]-(b)，否则 0

编辑：更正查询

Answer 1

所以，这是一种使用现有密码函数的方法。我不保证它的性能超级好，但试一试。我们正在使用 reduce 构建我们的集合，使用带有集合的累加器元组和我们查看的最后一个节点，因此我们可以检查它是否连接到下一个节点。这需要 2.0 的 case/when 语法——在 1.9 中可能有办法做到这一点，但它可能更复杂。

START a=node:node_auto_index(name="Trinity") 
MATCH a-[*]-b 
WHERE a <> b
WITH distinct a,b 
MATCH p = allshortestpaths(a-[*]-b) 
RETURN extract(x in nodes(p): x.name?), // a concise representation of the path we're checking
  head(
    reduce(acc=[[], head(nodes(p))], x IN tail(nodes(p)): // pop the first node off, traverse the tail
      CASE WHEN ALL (y IN tail(acc) WHERE y-->x)  // a bit of a hack because tail(acc)-->x doesn't parse right, so I had to wrap it so I can have a bare identifier in the pattern predicate
           THEN [head(acc) + 0, x]    // add a 0 to our accumulator collection
           ELSE [head(acc) + 1, x]    // add a 1 to our accumulator collection 
      END )) AS in_line

http://console.neo4j.org/r/v0jx03

输出：

+---------------------------------------------------------------------------+
| extract(x in nodes(p): x.name?)                               | in_line   |
+---------------------------------------------------------------------------+
| ["Trinity","Morpheus"]                                        | [1]       |
| ["Trinity","Morpheus","Cypher"]                               | [1,0]     |
| ["Trinity","Morpheus","Cypher","Agent Smith"]                 | [1,0,0]   |
| ["Trinity","Morpheus","Cypher","Agent Smith","The Architect"] | [1,0,0,0] |
| ["Trinity","Neo"]                                             | [1]       |
| ["Trinity","Neo",<null>]                                      | [1,1]     |
+---------------------------------------------------------------------------+

注意：感谢@boggle 的头脑风暴会议。