如何将neo4j节点的属性存储为数组? [英] How to store properties of a neo4j node as an array?

问题描述

我正在使用neo4j进行HR分析项目，遇到一个复杂的查询(我对cypher也很陌生).基本上，我列出了具有一些功能的员工列表，例如位置，技能集，教育程度以及具有位置，所需技能的职位.我查看位置(员工的位置和位置的位置)是否匹配，并将该分数设置为1(否则为0).如果我只有一个空缺职位，则查询工作正常.但是，随着位置的增加，变量将被最后一个位置的最后一个值覆盖.我可以想象将节点属性放在数组中，但是似乎密码没有(不推荐使用).

I am working on an HR analytics projects using neo4j and I encountered a complicated query (I am also very new to cypher). Basically, I have a list of employee with some features like location, skill set, education and positions with location, skills required. I look if location (employee’s location and position’s location) matches and set that score to 1 (0 otherwise). If I have only one open position, the query works fine. However, with more positions the variable gets overwritten with last value of last position. I would imagine putting the node properties in an array but doesn’t seem that cypher has that (deprecated).

https://neo4j.com/docs/rest-docs/current/#rest-api-property-values

这是我使用的查询，如果您能帮助我解决问题，将不胜感激.最终目标是在给定职位上为每位员工提供一个分数，并用得分最高的员工填补该职位.在cypher上甚至有可能吗? 非常感谢

This is the query that I use and I would be grateful if you can help me resolve the issue. Final goal is to have a score for every employee per a given position and fill the position with the employee with highest score. Is this even possible on cypher?? Thanks a lot

MATCH
  (e:Employee)-[r:FUTURE_POSITION]-> (p:Position {open_status:1}),
  (e)-[h:HAS_DEGREE]-> (d:Degree),
  (e)-[s:HAS_SKILL]-> (n:Personal_Skill)   
WITH r, e,p,d,n,
  CASE WHEN p.position_state = e.home_state THEN 1 ELSE 0 END AS SameStateScore,
  CASE WHEN p.position_city = e.home_city THEN 1 ELSE 0 END AS SameCityScore,
  CASE WHEN d.name = "College Degree" THEN 1 ELSE 0 END AS HasCollegeDegree,
  CASE WHEN n.name = "Management" THEN 1 ELSE 0 END AS HasRequiredSkill
SET e.score = SameStateScore + SameCityScore + HasCollegeDegree + HasRequiredSkill
RETURN DISTINCT e.name,p.name, SameStateScore,SameCityScore,HasCollegeDegree,MAX(HasRequiredSkill) AS HasRequiredSkill, e.score
ORDER BY e.score DESC

推荐答案

您在问题中使用的链接( HTTP API .

The link you used in your question (https://neo4j.com/docs/rest-docs/current/#rest-api-property-values) is for the deprecated legacy REST API. If you want to make HTTP requests to execute Cypher, you can use the new HTTP API instead.

为了每个职位都有单独的员工分数，您不应该将分数存储在Employee节点中，而应该存储在FUTURE_POSITION关系中-因为每个职位都有单独的关系.因此，只需使用r.score而不是e.score:

In order to have a separate employee score per position, you should not be storing the score in the Employee node, but in the FUTURE_POSITION relationship -- since there is a separate relationship per position. So, just use r.score instead of e.score:

MATCH
  (e:Employee)-[r:FUTURE_POSITION]->(p:Position {open_status:1}),
  (e)-[:HAS_DEGREE]-> (d:Degree),
  (e)-[:HAS_SKILL]-> (n:Personal_Skill)
WITH r, e, p,
  CASE WHEN p.position_state = e.home_state THEN 1 ELSE 0 END AS SameStateScore,
  CASE WHEN p.position_city = e.home_city THEN 1 ELSE 0 END AS SameCityScore,
  CASE WHEN d.name = "College Degree" THEN 1 ELSE 0 END AS HasCollegeDegree,
  CASE WHEN n.name = "Management" THEN 1 ELSE 0 END AS HasRequiredSkill
SET r.score = SameStateScore + SameCityScore + HasCollegeDegree + HasRequiredSkill
RETURN DISTINCT
  e.name, p.name, SameStateScore, SameCityScore, HasCollegeDegree,
  MAX(HasRequiredSkill) AS HasRequiredSkill, r.score
ORDER BY r.score DESC

此外，可能不需要DISTINCT选项，因为聚合函数(如MAX)已经确保同一结果行中的非聚合值集是不同的.

Also, the DISTINCT option may not be needed, as aggregation functions (like MAX) already make sure the set of non-aggregated values in the same result row are distinct.

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