Neo4j图形数据库中复杂匹配评分时的性能？

Question

我有Neo4J3.3.5图形数据库:27GB，50KK节点，500KK关系。索引打开。模式。个人电脑:16GB内存，4个核心。

MATCH (c:Company)-[r:HAS_BRANCH]->(b:Branch)
WHERE b.branchId in [27444, 1692, 23409, ...] //around 10 ids per query
RETURN 
c.companyId as Id, 
case r.branchType 
 when 0 then 25
 ... // //around 7 conditions per query 
 when 10 then 20 
end as Score

由于缺少后联合处理，我使用collect+unwind来合并所有关系的分数。

不幸的是，性能较低。我在5-10秒内得到一个关系（如上）查询的响应。当我试图将结果与collect+unwind组合时，查询“never”结束。

做这件事的更好/正确的方法是什么？也许我在图形设计上做错了什么？硬件配置到低？或者也许有一些算法可以与图库中的分数图（查询数据）匹配？

例如。用户可以搜索从西班牙生产木桌的新公司。

组合查询示例：

MATCH (c:Company)-[r:HAS_BRANCH]->(b:Branch)
WHERE b.branchId in ["27444" , "1692" , "23409" , "8744" , "9192" , "26591" , "21396" , "27151" , "20228" , "3517" , "25058" , "29549"] 
WITH case r.branchType 
when "0" then collect({id:c.companyId, score: 25}) 
 when "1" then collect({id:c.companyId, score: 19}) 
 when "2" then collect({id:c.companyId, score: 20}) 
 when "3" then collect({id:c.companyId, score: 19}) 
 when "4" then collect({id:c.companyId, score: 20}) 
 when "5" then collect({id:c.companyId, score: 15}) 
 when "6" then collect({id:c.companyId, score: 6}) 
 when "7" then collect({id:c.companyId, score: 5}) 
 when "8" then collect({id:c.companyId, score: 4}) 
 when "9" then collect({id:c.companyId, score: 4}) 
 when "10" then collect({id:c.companyId, score: 20}) 
end as rows
MATCH (c:Company)-[r:HAS_REVERTED_BRANCH]->(b:Branch)
WHERE b.branchId in ["27444" , "1692" , "23409" , "8744" , "9192" , "26591" , "21396" , "27151" , "20228" , "3517" , "25058" , "29549"] 
WITH rows + case r.branchType 
when "0" then collect({id:c.companyId, score: 25}) 
 when "1" then collect({id:c.companyId, score: 19}) 
 when "2" then collect({id:c.companyId, score: 20}) 
 when "3" then collect({id:c.companyId, score: 19}) 
 when "10" then collect({id:c.companyId, score: 20}) 
end as rows
MATCH (c:Company)-[r:HAS_COUNTRY]->(cou:Country)
WHERE cou.countryId in ["9580" , "18551" , "15895"] 
WITH rows + case r.branchType 
when "0" then collect({id:c.companyId, score: 30}) 
 when "2" then collect({id:c.companyId, score: 15}) 
 end as rows
... //here I would add in future other relations scoring
UNWIND rows AS row
RETURN row.id AS Id, sum(row.score) AS Score
ORDER BY Score DESC
LIMIT 100

Answer 1

您可以尝试这个查询，看看它是否更好：

MATCH (c:Company) WITH c
OPTIONAL MATCH (c)-[r1:HAS_BRANCH]->(b:Branch) WHERE b.branchId in ["27444" , "1692" , "23409" , "8744" , "9192" , "26591" , "21396" , "27151" , "20228" , "3517" , "25058" , "29549"] 
OPTIONAL MATCH (c)-[r2:HAS_REVERTED_BRANCH]->(c:Branch) WHERE c.branchId in ["27444" , "1692" , "23409" , "8744" , "9192" , "26591" , "21396" , "27151" , "20228" , "3517" , "25058" , "29549"] 
OPTIONAL MATCH (c)-[r3:HAS_COUNTRY]->(cou:Country) WHERE cou.countryId in ["9580" , "18551" , "15895"] 
WITH c, 
    case r1.branchType 
      when "0" then 25
      when "1" then 19 
      when "2" then 20 
      when "3" then 19 
      when "4" then 20 
      when "5" then 15 
      when "6" then 6 
      when "7" then 5 
      when "8" then 4 
      when "9" then 4 
      when "10" then 20 
    end as branchScore,
    case r2.branchType 
      when "0" then  25 
      when "1" then  19 
      when "2" then  20 
      when "3" then  19 
      when "10" then  20 
    end as revertedBranchScore,
    case r3.branchType 
      when "0" then  30
      when "2" then  15 
    end as countryScore

WITH c.id AS Id, branchScore + revertedBranchScore + countryScore AS Score
RETURN Id, sum(Score) AS Score
ORDER BY Score DESC
LIMIT 100

或者更好的方法是这个（但前提是公司节点必须链接到国家和分支):

MATCH 
  (c:Company)-[r1:HAS_BRANCH]->(b:Branch),
  (c)-[r2:HAS_REVERTED_BRANCH]->(c:Branch),
  (c)-[r3:HAS_COUNTRY]->(cou:Country)
WHERE 
  b.branchId in ["27444" , "1692" , "23409" , "8744" , "9192" , "26591" , "21396" , "27151" , "20228" , "3517" , "25058" , "29549"] AND 
  c.branchId in ["27444" , "1692" , "23409" , "8744" , "9192" , "26591" , "21396" , "27151" , "20228" , "3517" , "25058" , "29549"] AND
  cou.countryId in ["9580" , "18551" , "15895"]
WITH c, 
    case r1.branchType 
      when "0" then 25
      when "1" then 19 
      when "2" then 20 
      when "3" then 19 
      when "4" then 20 
      when "5" then 15 
      when "6" then 6 
      when "7" then 5 
      when "8" then 4 
      when "9" then 4 
      when "10" then 20 
    end as branchScore,
    case r2.branchType 
      when "0" then  25 
      when "1" then  19 
      when "2" then  20 
      when "3" then  19 
      when "10" then  20 
    end as revertedBranchScore,
    case r3.branchType 
      when "0" then  30
      when "2" then  15 
    end as countryScore

WITH c.id AS Id, branchScore + revertedBranchScore + countryScore AS Score
RETURN Id, sum(Score) AS Score
ORDER BY Score DESC
LIMIT 100