Question

我在这里向你求助。我有一个neo4j数据库，有完整的k-ary树，有数百万个节点。节点和边缘都具有恒定数量的属性（可能从节点到边缘不同，但所有节点都具有完全x属性，所有边缘都具有y属性）。我的任务是返回节点和边缘上所有属性的总和。我试过这个问题：

Match p=(:Vertex_ss3 {name:'vertex_1594320'})-[:EDGE_ss3*]->(:Vertex_ss3 {name:'vertex_1'}) 
return 
reduce(sum = 0, n IN nodes(p) | sum + n.attr1) as tot_attr1_node,
reduce(sum = 0, n IN nodes(p) | sum + n.attr2) as tot_attr2_node, 
reduce(sum = 0, n IN nodes(p) | sum + n.attr3) as tot_attr3_node,
reduce(sum = 0, n IN nodes(p) | sum + n.attr4) as tot_attr4_node, 
reduce(sum = 0, n IN nodes(p) | sum + n.attr5) as tot_attr5_node,
reduce(sum = 0, n IN nodes(p) | sum + n.attr6) as tot_attr6_node,
reduce(sum = 0, n IN relationships(p) | sum + n.attr1) as tot_attr1_edge,
reduce(sum = 0, n IN relationships(p) | sum + n.attr2) as tot_attr2_edge,
reduce(sum = 0, n IN relationships(p) | sum + n.attr3) as tot_attr3_edge,
reduce(sum = 0, n IN relationships(p) | sum + n.attr4) as tot_attr4_edge,
reduce(sum = 0, n IN relationships(p) | sum + n.attr5) as tot_attr5_edge

使用深度为13的3-ary树返回大约需要13/14秒。有没有办法在时间上有所改进？

我真的不知道节点（p）和关系（p）是如何工作的，但是当我编写查询时，似乎对于每个属性，数据库必须从路径中检索所有节点或所有关系，isn'有没有办法一劳永逸地做到这一点？

感谢您的建议：）

Answer 1

您应该能够使用可以并行执行查询（具有不同数据）的APOC procedures之一。不幸的是，它们的记录很少（有些根本没有记录）。因此，我将提供一个关于其中一个程序的半教程，以及它如何帮助您获得更快的结果。

下面的查询是如何使用apoc.cypher.mapParallel并行地对每个节点属性求和（至少可能;该过程确定实际的并行度），然后并行地对每个关系属性求和。

MATCH p=(:Vertex_ss3 {name:'vertex_1594320'})-[:EDGE_ss3*]->(:Vertex_ss3 {name:'vertex_1'}) 
CALL apoc.cypher.mapParallel(
  'UNWIND nodes AS n RETURN _ AS attr, SUM(n[_]) AS sum',
  {nodes: NODES(p)},
  ['attr1','attr2','attr3','attr4','attr5','attr6']) YIELD value AS nodeAttr
WITH p, COLLECT(nodeAttr) AS nodeAttrs
CALL apoc.cypher.mapParallel(
  'UNWIND rels AS n RETURN _ AS attr, SUM(n[_]) AS sum',
  {rels: RELATIONSHIPS(p)},
  ['attr1','attr2','attr3','attr4','attr5']) YIELD value AS relAttr
RETURN nodeAttrs, COLLECT(relAttr) AS relAttrs;

该过程的第一个参数是您要并行运行的Cypher查询。
该过程的第二个参数定义传递给过程的Cypher查询的参数。对于每个参数，该过程创建一个具有相同名称的标识符（例如，＆＃34; {foo}＆＃34;的值只能通过使用foo标识符来访问。）
< / LI>
该过程确保下划线（＆＃34; _＆＃34;）标识符将具有传递给过程的列表中的一个元素（最后一个参数）的值。
< / LI>
nodeAttrs和relAttrs将是{attr：...，sum：...}地图的集合。

Answer 2

不确定这是更快还是更慢，但您可以尝试展开节点和关系，然后使用SUM（）函数获取所需的值。

MATCH p=(:Vertex_ss3 {name:'vertex_1594320'})-[:EDGE_ss3*]->(:Vertex_ss3 {name:'vertex_1'}) 
WITH p
UNWIND nodes(p) as n
WITH p, 
SUM(n.attr1) as tot_attr1_node
SUM(n.attr2) as tot_attr2_node
SUM(n.attr3) as tot_attr3_node
SUM(n.attr4) as tot_attr4_node
SUM(n.attr5) as tot_attr5_node
SUM(n.attr6) as tot_attr6_node
UNWIND relationships(p) as n
WITH p, tot_attr1_node, tot_attr2_node, tot_attr3_node, tot_attr4_node, tot_attr5_node, tot_attr6_node
SUM(n.attr1) as tot_attr1_edge
SUM(n.attr2) as tot_attr2_edge
SUM(n.attr3) as tot_attr3_edge
SUM(n.attr4) as tot_attr4_edge
SUM(n.attr5) as tot_attr5_edge
RETURN tot_attr1_node, 
tot_attr2_node, 
tot_attr3_node, 
tot_attr4_node, 
tot_attr5_node, 
tot_attr6_node, 
tot_attr1_edge, 
tot_attr2_edge, 
tot_attr3_edge, 
tot_attr4_edge, 
tot_attr5_edge

总结所有属性Neo4j

2 个答案: