neo4j cypher基于属性字符串值

Question

我有一个带标签实验的节点，其中有一个名为ExperimentName的属性。

此ExperimentName基于3个不同变量的串联

“条件（ExperimentProfile1）剂量”

示例：

Control diet (MOA77) LD
Control gavage(MOA66) HD

我有另一个名为ExperimentMapper的节点 它有3个属性： - 条件 - ExperimentProfile - 剂量

当experimentName是3个属性的结果时，我想在Node Experiment 和Node ExperimentMapper 之间创建一个关系。

我曾尝试使用Regex，但查询速度非常慢并且需要永远...

任何帮助？

这是我的密码，但尽管我创建了索引

，但它仍在继续

MATCH (mxpExperiment:MxpExperiment) OPTIONAL MATCH (otuExperimentMapper:OtuExperimentMapper) 
WHERE  mxpExperiment.name CONTAINS otuExperimentMapper.Condition 
AND mxpExperiment.name CONTAINS otuExperimentMapper.Experiment
AND mxpExperiment.name CONTAINS otuExperimentMapper.dose

CREATE (mxpExperiment)-[:OTU_EXPERIMENT_MAPPER]->(otuExperimentMapper)
RETURN mxpExperiment, otuExperimentMapper

Answer 1

我认为您需要从Experiment Mapper。

的一侧开始

首先，您需要创建一个索引：

CREATE INDEX ON :MxpExperiment(name)

然后查询可以如下：

MATCH (otuExperimentMapper:OtuExperimentMapper) 
WITH otuExperimentMapper,
     otuExperimentMapper.Condition  + ' (' +
     otuExperimentMapper.Experiment + ') ' +
     otuExperimentMapper.dose AS name
MATCH (mxpExperiment:MxpExperiment) WHERE mxpExperiment.name = name
MERGE (mxpExperiment)-[:OTU_EXPERIMENT_MAPPER]->(otuExperimentMapper)
RETURN mxpExperiment, otuExperimentMapper