sparql - 检索每个 OWL 的 unionOf 和 intersectionOf 的集合

Question

我正在尝试在 OWL 文件中提取intersectionOf和，该文件由类集合或/和. 我创建了一个 SPARQL 查询，它为提取“集合” ，但问题是某些检索到的数据与该类无关。unionOfinteresctionOfunionOfsomeValuesFromonPropertyintersectionOf

例如，我有一个名为man. 这个类有一个等价类，它是三个intersectionOf类，即、adult和person。我的maleSPARQL 查询返回一些不正确的结果：它返回类adult、person和也与我的 OWL 文件中的所有其他类等效的类，例如，这是不正确的。这是我的 SPARQL 查询：malemanhaulage_worker

PREFIX abc: <http://owl.cs.manchester.ac.uk/2009/07/sssw/people#>
PREFIX ghi: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX mno: <http://www.w3.org/2001/XMLSchema#>
PREFIX owl: <http://www.w3.org/2002/07/owl#>
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX list: <http://jena.hpl.hp.com/ARQ/list#>

SELECT  Distinct ?class ?equivalentClass  
WHERE{ ?class a owl:Class .
FILTER( STRSTARTS(STR(?class),"http://www.w3.org/2002/07/owl#") || STRSTARTS(STR(?class),"http://owl.cs.manchester.ac.uk/2009/07/sssw/people#") 

)

     ?x a owl:Class ; owl:intersectionOf ?list .
     ?list rdf:rest*/rdf:first ?equivalentClass .


   } GROUP BY ?class ?equivalentClass ORDER BY ?no

这是我的 OWL 文件：

<?xml version="1.0"?>
<rdf:RDF
    xmlns="http://owl.cs.manchester.ac.uk/2009/07/sssw/people#"
    xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
    xmlns:xsd="http://www.w3.org/2001/XMLSchema#"
    xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#"
    xmlns:owl="http://www.w3.org/2002/07/owl#"
    xmlns:ns0="http://owl.cs.manchester.ac.uk/2009/07/sssw/people#"
  xml:base="http://owl.cs.manchester.ac.uk/2009/07/sssw/people">
  <owl:Ontology rdf:about="http://owl.cs.manchester.ac.uk/2009/07/sssw/people"/>
  <owl:Class rdf:about="http://www.w3.org/2002/07/owl#Thing"/>
  <owl:Class rdf:about="http://owl.cs.manchester.ac.uk/2009/07/sssw/people#haulage_worker">
    <rdfs:comment rdf:datatype="http://www.w3.org/2001/XMLSchema#string"
   ></rdfs:comment>
    <owl:equivalentClass>
      <owl:Restriction>
        <owl:onProperty>
          <owl:ObjectProperty      rdf:about="http://owl.cs.manchester.ac.uk/2009/07/sssw/people#works_for"/>
        </owl:onProperty>
        <owl:someValuesFrom>
        <owl:Class>
            <owl:unionOf rdf:parseType="Collection">
              <owl:Restriction>
               <owl:onProperty>
                <owl:ObjectProperty    rdf:about="http://owl.cs.manchester.ac.uk/2009/07/sssw/people#part_of"/>
               </owl:onProperty>
               <owl:someValuesFrom>
                <owl:Class   rdf:about="http://owl.cs.manchester.ac.uk/2009/07/sssw/people#haulage_company"/>
               </owl:someValuesFrom>
              </owl:Restriction>
              <owl:Class   rdf:about="http://owl.cs.manchester.ac.uk/2009/07/sssw/people#haulage_company"/>
            </owl:unionOf>
          </owl:Class>
         </owl:someValuesFrom>
        </owl:Restriction>
     </owl:equivalentClass>
    <rdfs:label rdf:datatype="http://www.w3.org/2001/XMLSchema#string"
    >haulage worker</rdfs:label>
  </owl:Class>
  <owl:Class rdf:about="http://owl.cs.manchester.ac.uk/2009/07/sssw/people#man">
    <owl:equivalentClass>
      <owl:Class>
        <owl:intersectionOf rdf:parseType="Collection">
          <owl:Class rdf:about="http://owl.cs.manchester.ac.uk/2009/07/sssw/people#adult"/>
          <owl:Class rdf:about="http://owl.cs.manchester.ac.uk/2009/07/sssw/people#person"/>
          <owl:Class rdf:about="http://owl.cs.manchester.ac.uk/2009/07/sssw/people#male"/>
        </owl:intersectionOf>
      </owl:Class>
    </owl:equivalentClass>
  </owl:Class>
</rdf:RDF>

这是我得到的输出（它们不是正确的输出）：

-----------------------------------------
    | class               | equivalentClass |
    =========================================
    | abc:adult           | abc:adult       |
    | abc:adult           | abc:male        |
    | abc:adult           | abc:person      |
    | abc:haulage_company | abc:adult       |
    | abc:haulage_company | abc:male        |
    | abc:haulage_company | abc:person      |
    | abc:haulage_worker  | abc:adult       |
    | abc:haulage_worker  | abc:male        |
    | abc:haulage_worker  | abc:person      |
    | abc:male            | abc:adult       |
    | abc:male            | abc:male        |
    | abc:male            | abc:person      |
    | abc:man             | abc:adult       |
    | abc:man             | abc:male        |
    | abc:man             | abc:person      |
    | abc:person          | abc:adult       |
    | abc:person          | abc:male        |
    | abc:person          | abc:person      |
    | owl:Thing           | abc:adult       |
    | owl:Thing           | abc:male        |
    | owl:Thing           | abc:person      |
    -----------------------------------------

预期的输出将是这样的：

-----------------------------------------
    | class               | equivalentClass |
    =========================================
    | abc:adult           | abc:adult       |
    | abc:adult           | abc:male        |
    | abc:adult           | abc:person      |
    | abc:haulage_company |                 |
    | abc:haulage_company |                 |
    | abc:haulage_company |                 |
    | abc:haulage_worker  |                 |
    | abc:haulage_worker  |                 |
    | abc:haulage_worker  |                 |
    | abc:male            | abc:adult       |
    | abc:male            | abc:male        |
    | abc:male            | abc:person      |
    | abc:man             | abc:adult       |
    | abc:man             | abc:male        |
    | abc:man             | abc:person      |
    | abc:person          | abc:adult       |
    | abc:person          | abc:male        |
    | abc:person          | abc:person      |
    | owl:Thing           |                 |
    | owl:Thing           |                 |
    | owl:Thing           |                 |
    -----------------------------------------

我应该在我的 SPARQL 查询中进行哪些更改以使我的输出像上一个表一样？

score 2 · Accepted Answer

稍微清理一下您的查询，我们有：

prefix abc: <http://owl.cs.manchester.ac.uk/2009/07/sssw/people#>
prefix ghi: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
prefix mno: <http://www.w3.org/2001/XMLSchema#>
prefix owl: <http://www.w3.org/2002/07/owl#>
prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#>
prefix rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
prefix list: <http://jena.hpl.hp.com/ARQ/list#>

select distinct ?class ?equivalentClass where {
  ?class a owl:Class .
  filter( strstarts(str(?class),str(owl:)) ||     # since "owl:" is an IRI, you can 
          strstarts(str(?class),str(abc:))    )   # use str(owl:) and str(:abc)

  ?x a owl:Class ; 
     owl:intersectionOf ?list .
     ?list rdf:rest*/rdf:first ?equivalentClass .
}
group by ?class ?equivalentClass
order by ?class                                   # ?class, not ?no

您的问题在于您正在选择?class，它可以是owl:Class本体中的每个（只要它以适当的前缀开头），然后?equivalentClass从的交集类列表中选择?x，并且与?no ?class?x没有任何联系`。）?class. (You were also sorting by, but I think you meant to sort by

如果我们以更易于阅读的格式（例如 Turtle）查看数据，找出要编写的正确查询会更容易。在 Turtle 中，man类是：

ns0:man  a                   owl:Class ;
        owl:equivalentClass  [ a                   owl:Class ;
                               owl:intersectionOf  ( ns0:adult ns0:person ns0:male )
                             ] .

您正在寻找的东西是owl:Classes，与owl:equivalentClass其他东西相关，它是owl:Class，并且具有的列表值owl:intersectionOf。这在 SPARQL 中并不太难，而且查询实际上与这个 Turtle 文本具有相同的结构：

prefix abc: <http://owl.cs.manchester.ac.uk/2009/07/sssw/people#>
prefix ghi: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
prefix mno: <http://www.w3.org/2001/XMLSchema#>
prefix owl: <http://www.w3.org/2002/07/owl#>
prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#>
prefix rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
prefix list: <http://jena.hpl.hp.com/ARQ/list#>

select distinct ?class ?otherClass where {
  ?class a owl:Class ;
         owl:equivalentClass [ a owl:Class ;
                               owl:intersectionOf [ rdf:rest*/rdf:first ?otherClass ] ] .
  filter( strstarts(str(?class),str(owl:)) ||
          strstarts(str(?class),str(abc:))    )
}
group by ?class ?otherClass
order by ?class

我将变量名从更改equivalentClass为otherClass，因为adult,male和person不等同于man. 他们的交点是。使用 Jena 的命令行sparql工具，您将获得如下结果：

$ sparql --data data.rdf --query query.rq
------------------------
| class   | otherClass |
========================
| abc:man | abc:adult  |
| abc:man | abc:male   |
| abc:man | abc:person |
------------------------

此查询仅检索与某个交集等效的类。您的预期结果显示了 IRI 以abc:or开头的所有类owl:，这意味着额外的结构实际上是可选的，因此我们通过将可选部分包装在中相应地调整查询optional { … }，我们得到了我们正在寻找的结果：

prefix abc: <http://owl.cs.manchester.ac.uk/2009/07/sssw/people#>
prefix ghi: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
prefix mno: <http://www.w3.org/2001/XMLSchema#>
prefix owl: <http://www.w3.org/2002/07/owl#>
prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#>
prefix rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
prefix list: <http://jena.hpl.hp.com/ARQ/list#>

select distinct ?class ?otherClass where {
  ?class a owl:Class .
  optional { 
    ?class owl:equivalentClass [ a owl:Class ;
                                 owl:intersectionOf [ rdf:rest*/rdf:first ?otherClass ] ] .
  }
  filter( strstarts(str(?class),str(owl:)) ||
          strstarts(str(?class),str(abc:))    )
}
group by ?class ?otherClass
order by ?class

$ sparql --data data.rdf --query query.rq
------------------------------------
| class               | otherClass |
====================================
| abc:adult           |            |
| abc:haulage_company |            |
| abc:haulage_worker  |            |
| abc:male            |            |
| abc:man             | abc:adult  |
| abc:man             | abc:male   |
| abc:man             | abc:person |
| abc:person          |            |
| owl:Thing           |            |
------------------------------------

sparql - 检索每个 OWL 的 unionOf 和 intersectionOf 的集合

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