Neo4j多MATCH语句返回重复值问题求助

现有总计6个节点，分为time1和time2两个标签，每个标签下各3个节点。执行：

match (a:time1) return collect(a)

可得到预期的3个time1节点列表；但执行：

match (a:time1), (b:time2) return collect(a), collect(b)

时，预期得到各含3个元素的两个列表，实际却得到各含9个元素且每个元素重复3次的列表，输出CSV内容如下：

collect(a),collect(b)
"[(:time1 {con1: 0.9,con2: 0.0,con3: 0.0}), (:time1 {con1: 0.9,con2: 0.0,con3: 0.0}), (:time1 {con1: 0.9,con2: 0.0,con3: 0.0}), (:time1 {con1: 0.0,con2: 0.4,con3: 0.0}), (:time1 {con1: 0.0,con2: 0.4,con3: 0.0}), (:time1 {con1: 0.0,con2: 0.4,con3: 0.0}), (:time1 {con1: 0.0,con2: 0.0,con3: 0.5}), (:time1 {con1: 0.0,con2: 0.0,con3: 0.5}), (:time1 {con1: 0.0,con2: 0.0,con3: 0.5})]","[(:time2 {con1: 0.9,con2: 0.0,con3: 0.0}), (:time2 {con1: 0.0,con2: 0.4,con3: 0.0}), (:time2 {con1: 0.0,con2: 0.0,con3: 0.5}), (:time2 {con1: 0.9,con2: 0.0,con3: 0.0}), (:time2 {con1: 0.0,con2: 0.4,con3: 0.0}), (:time2 {con1: 0.0,con2: 0.0,con3: 0.5}), (:time2 {con1: 0.9,con2: 0.0,con3: 0.0}), (:time2 {con1: 0.0,con2: 0.4,con3: 0.0}), (:time2 {con1: 0.0,con2: 0.0,con3: 0.5})]"

原因分析与解决方案 #

原因分析 #

MATCH (a:time1), (b:time2)会生成笛卡尔积——每个time1节点会和每个time2节点配对，3个time1×3个time2=9条结果行。当你用collect(a)和collect(b)时，会把这9行里的所有a和b分别收集起来，所以每个time1节点会重复3次（对应3个time2节点的配对），每个time2节点也会重复3次，最终得到各含9个元素的列表。

解决方案 #

要避免笛卡尔积，分别收集两个标签的节点即可，推荐以下几种写法：

方法1：分步骤收集（高效推荐）

先用WITH收集其中一个标签的节点，再收集另一个，从根源避免笛卡尔积：

MATCH (a:time1)
WITH collect(a) AS time1Nodes
MATCH (b:time2)
RETURN time1Nodes, collect(b) AS time2Nodes

方法2：使用DISTINCT去重

如果一定要在同一个MATCH中编写，可在collect中加入DISTINCT去重，但这种方法仍会先生成笛卡尔积再去重，数据量大时效率较低：

MATCH (a:time1), (b:time2)
RETURN collect(DISTINCT a) AS time1Nodes, collect(DISTINCT b) AS time2Nodes

方法3：子查询分别获取集合

通过子查询独立获取两个标签的节点集合：

RETURN
  (MATCH (a:time1) RETURN collect(a) AS time1Nodes).time1Nodes,
  (MATCH (b:time2) RETURN collect(b) AS time2Nodes).time2Nodes

内容的提问来源于stack exchange，提问作者Novabro