Neo4j中的sparql基准查询速度慢

我正在尝试使用neo4j实现SPARQL查询。我已经使用三元组创建了Neo4j图

为了总结数据加载，我的图表有以下结构：

Subject   => Node
Predicate => Relationship
Object    => Node 

若谓词是日期、字符串、整数原语，那个么将创建一个属性而不是关系，并存储在节点中

现在，我正在尝试以下在Noe4j中非常慢的查询

Query 4: Feature with the highest ratio between price with that feature and price without that feature. 

    corresponding SPARQL query for this, 

            prefix bsbm: <http://www4.wiwiss.fu-berlin.de/bizer/bsbm/v01/vocabulary/>
            prefix bsbm-inst: <http://www4.wiwiss.fu-berlin.de/bizer/bsbm/v01/instances/>
            prefix xsd: <http://www.w3.org/2001/XMLSchema#>

            Select ?feature ((?sumF*(?countTotal-?countF))/(?countF*(?sumTotal-?sumF)) As ?priceRatio)
            {
              { Select (count(?price) As ?countTotal) (sum(xsd:float(str(?price))) As ?sumTotal)
                {
                  ?product a <http://www4.wiwiss.fu-berlin.de/bizer/bsbm/v01/instances/ProductType294> .
                  ?offer bsbm:product ?product ;
                         bsbm:price ?price .
                }
              }
              { Select ?feature (count(?price2) As ?countF) (sum(xsd:float(str(?price2))) As ?sumF)
                {
                  ?product2 a <http://www4.wiwiss.fu-berlin.de/bizer/bsbm/v01/instances/ProductType294> ;
                           bsbm:productFeature ?feature .
                  ?offer2 bsbm:product ?product2 ;
                         bsbm:price ?price2 .
                }
                Group By ?feature
              }
            }
           Order By desc(?priceRatio) ?feature
           Limit 100
 Cypher query I created for this,

    MATCH p1 = (offer1:Offer)-[r1:`product`]->(products1:ProductType294)
    MATCH p2 = (offer2:Offer)-[r2:`product`]->products2:ProductType294)-[:`productFeature`]->features
    return (sum( DISTINCT offer2.price) * ( count( DISTINCT offer1.price) - count( DISTINCT offer2.price)) /(count(DISTINCT offer2.price)*(sum( DISTINCT offer1.price) - sum(DISTINCT offer2.price)))) AS cnt,features.__URI__ AS frui
    ORDER BY cnt DESC,frui 

这个查询真的很慢，请让我知道我是否以错误的方式制定了查询

Another query is Query 5: Show the most popular products of a specific product type for each country - by review count ,

      prefix bsbm: <http://www4.wiwiss.fu-berlin.de/bizer/bsbm/v01/vocabulary/>
      prefix bsbm-inst: <http://www4.wiwiss.fu-berlin.de/bizer/bsbm/v01/instances/>
      prefix rev: <http://purl.org/stuff/rev#>
      prefix xsd: <http://www.w3.org/2001/XMLSchema#>

      Select ?country ?product ?nrOfReviews ?avgPrice
      {
        { Select ?country (max(?nrOfReviews) As ?maxReviews)
          {
            { Select ?country ?product (count(?review) As ?nrOfReviews)
              {
                ?product a <http://www4.wiwiss.fu-berlin.de/bizer/bsbm/v01/instances/ProductType403> .
                ?review bsbm:reviewFor ?product ;
                        rev:reviewer ?reviewer .
                ?reviewer bsbm:country ?country .
              }
              Group By ?country ?product
            }
          }
          Group By ?country
        }
        { Select ?product (avg(xsd:float(str(?price))) As ?avgPrice)
          {
            ?product a <http://www4.wiwiss.fu-berlin.de/bizer/bsbm/v01/instances/ProductType403> .
            ?offer bsbm:product ?product .
            ?offer bsbm:price ?price .
          }
          Group By ?product
        }
        { Select ?country ?product (count(?review) As ?nrOfReviews)
          {
            ?product a <http://www4.wiwiss.fu-berlin.de/bizer/bsbm/v01/instances/ProductType403> .
            ?review bsbm:reviewFor ?product .
            ?review rev:reviewer ?reviewer .
            ?reviewer bsbm:country ?country .
          }
          Group By ?country ?product
        }
        FILTER(?nrOfReviews=?maxReviews)
      }
      Order By desc(?nrOfReviews) ?country ?product

Cypher query I created for this is following,

    MATCH (products2:ProductType403)<-[:`reviewFor`]-(reviews:Review)-[:`reviewer`]->(rvrs)-[:`country`]->(countries)
    with count(reviews) AS reviewcount,products2.__URI__ AS pruis, countries.__URI__ AS cntrs
    MATCH (products1:ProductType403)<-[:`product`]-(offer:Offer)
    with AVG(offer.price) AS avgPrice, MAX(reviewcount) AS maxrevs, cntrs
    MATCH (products2:ProductType403)<-[:`reviewFor`]-(reviews:Review)-[:`reviewer`]->(rvrs)-[:`country`]->(countries)
    with avgPrice, maxrevs,countries, count(reviews) AS rvs, countries.__URI__ AS curis, products2.__URI__ AS puris
    where maxrevs=rvs
    RETURN curis,puris,rvs,avgPrice

甚至这个查询也非常慢。我是否以正确的方式表述查询

我有1000万个三倍的柏林基准数据集 每个类型谓词都转换为标签。 对于查询4，我试图得到的是价格与价格之间的比率最高的特性 该功能和没有该功能的价格。这是去巴黎的路吗 制定查询？ 对于查询4，我得到了此查询的正确结果。 如果我不计算总和和计数，那么查询就会很快执行。

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提前感谢：SPARQL查询和信息可以在以下位置找到：

对我来说，这些看起来像全局图查询？ 数据集的大小是多少

在两条路径之间创建笛卡尔积？ 这两条路不应该有某种联系吗

ProductType标签上不应该有属性类型吗产品类型{type:294} 如果有的话，你会在：ProductTypetype和：OrderNo上有一个索引

我真的不明白这个计算

计算不同价格的增量乘以报价的不同价格之和2 通过 报价的不同价格计数2，乘以两个订单价格之和的增量

MATCH (offer1:Offer)-[r1:`product`]->(products1:ProductType294)
MATCH (offer2:Offer)-[r2:`product`]->(products2:ProductType294)-[:`productFeature`]->features

RETURN (sum( DISTINCT offer2.price) * 
       ( count( DISTINCT offer1.price) - count( DISTINCT offer2.price)) 
       / (count(DISTINCT offer2.price)*
       (sum( DISTINCT offer1.price) - sum(DISTINCT offer2.price)))) 
       AS cnt,features.__URI__ AS frui
ORDER BY cnt DESC,frui 

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也许您可以共享导入的数据库，并添加正在使用的图形模型的模型图片。那我们可以试着帮你。仅仅从描述来看，我真的不明白你在做什么。除了您的查询似乎是图形全局查询之外。