检索OWL交集类隐含的超类

Question

OWL本体可以具有类A,B和C以及公理(在DL表示法中):

A⊑(B⊓C)

或者近似曼彻斯特OWL语法:

甲subClassOf(B 和 C)

从逻辑上讲,A是B的子类,A是C的子类,但是三元组

A rdfs:subClassOf B
A rdfs:subClassOf C

不一定存在于OWL本体的RDF序列化中.例如,考虑Protégé中这个非常简单的本体及其在RDF/XML和Turtle中的RDF序列化:

<rdf:RDF
    xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
    xmlns="http://stackoverflow.com/q/19924861/1281433/sample.owl#"
    xmlns:owl="http://www.w3.org/2002/07/owl#"
    xmlns:xsd="http://www.w3.org/2001/XMLSchema#"
    xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#">
  <owl:Ontology rdf:about="http://stackoverflow.com/q/19924861/1281433/sample.owl"/>
  <owl:Class rdf:about="http://stackoverflow.com/q/19924861/1281433/sample.owl#C"/>
  <owl:Class rdf:about="http://stackoverflow.com/q/19924861/1281433/sample.owl#B"/>
  <owl:Class rdf:about="http://stackoverflow.com/q/19924861/1281433/sample.owl#A">
    <rdfs:subClassOf>
      <owl:Class>
        <owl:intersectionOf rdf:parseType="Collection">
          <owl:Class rdf:about="http://stackoverflow.com/q/19924861/1281433/sample.owl#B"/>
          <owl:Class rdf:about="http://stackoverflow.com/q/19924861/1281433/sample.owl#C"/>
        </owl:intersectionOf>
      </owl:Class>
    </rdfs:subClassOf>
  </owl:Class>
</rdf:RDF>

@prefix :      <http://stackoverflow.com/q/19924861/1281433/sample.owl#> .
@prefix rdfs:  <http://www.w3.org/2000/01/rdf-schema#> .
@prefix owl:   <http://www.w3.org/2002/07/owl#> .
@prefix xsd:   <http://www.w3.org/2001/XMLSchema#> .
@prefix rdf:   <http://www.w3.org/1999/02/22-rdf-syntax-ns#> .

<http://stackoverflow.com/q/19924861/1281433/sample.owl>
        a       owl:Ontology .

:B      a       owl:Class .

:C      a       owl:Class .

:A      a                owl:Class ;
        rdfs:subClassOf  [ a                   owl:Class ;
                           owl:intersectionOf  ( :B :C )
                         ] .

序列化有一个三元组rdfs:subClassOf,但对象不是:B或者:C,所以查询就像

:A rdfs:subClassOf ?superclass

不会返回超类:A.如何编写将返回那些超类的SPARQL查询:A？

Answer 1

听起来你有一个类是某个交集类的子类.例如,你可能有

学生 ⊑ 人 ⊓ 在某些课程中注册

在ProtégéOWL本体编辑器中,这看起来像:

如果为子类编写SPARQL查询,例如,

prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#>

select ?subclass ?superclass where { 
  ?subclass rdfs:subClassOf ?superclass
}

并且您没有推理器推断其他数据,您不会在结果中看到Student作为子类,但您可能会看到一个空白(匿名)节点:

---------------------------------------------------------
| subclass                                 | superclass |
=========================================================
| <http://www.examples.org/school#Student> | _:b0       |
---------------------------------------------------------

要理解为什么会出现这种情况,您需要查看本体的RDF序列化.在这种情况下,它是(在RDF/XML中):

<rdf:RDF
    xmlns="http://www.examples.org/school#"
    xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
    xmlns:owl="http://www.w3.org/2002/07/owl#"
    xmlns:xsd="http://www.w3.org/2001/XMLSchema#"
    xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#">
  <owl:Ontology rdf:about="http://www.examples.org/school"/>
  <owl:Class rdf:about="http://www.examples.org/school#Course"/>
  <owl:Class rdf:about="http://www.examples.org/school#Person"/>
  <owl:Class rdf:about="http://www.examples.org/school#Student">
    <rdfs:subClassOf>
      <owl:Class>
        <owl:intersectionOf rdf:parseType="Collection">
          <owl:Class rdf:about="http://www.examples.org/school#Person"/>
          <owl:Restriction>
            <owl:onProperty>
              <owl:ObjectProperty rdf:about="http://www.examples.org/school#enrolledIn"/>
            </owl:onProperty>
            <owl:someValuesFrom rdf:resource="http://www.examples.org/school#Course"/>
          </owl:Restriction>
        </owl:intersectionOf>
      </owl:Class>
    </rdfs:subClassOf>
  </owl:Class>
</rdf:RDF>

或者在更易读的Turtle中(也更像是SPARQL查询语法):

@prefix :      <http://www.examples.org/school#> .
@prefix rdfs:  <http://www.w3.org/2000/01/rdf-schema#> .
@prefix owl:   <http://www.w3.org/2002/07/owl#> .
@prefix xsd:   <http://www.w3.org/2001/XMLSchema#> .
@prefix rdf:   <http://www.w3.org/1999/02/22-rdf-syntax-ns#> .

:Student  a              owl:Class ;
        rdfs:subClassOf  [ a                   owl:Class ;
                           owl:intersectionOf  ( :Person [ a                   owl:Restriction ;
                                                           owl:onProperty      :enrolledIn ;
                                                           owl:someValuesFrom  :Course
                                                         ] )
                         ] .

:Person  a      owl:Class .

:enrolledIn  a  owl:ObjectProperty .

:Course  a      owl:Class .

<http://www.examples.org/school>
        a       owl:Ontology .

事实上,Student rdfs:subClassOf [ ... ]数据中有三个,但它[ ... ]是一个空白节点; 它是一个匿名的owl:Class,是其他一些类的交集.推理机将能够告诉你,如果X ⊑(Ÿ和Z ^),那么X ⊑ Ÿ 和X ⊑ ž,但本身就是一个SPARQL查询不会那样做.你可以制作一个更复杂的SPARQL查询,但是:

prefix rdfs:  <http://www.w3.org/2000/01/rdf-schema#>
prefix owl:   <http://www.w3.org/2002/07/owl#>
prefix rdf:   <http://www.w3.org/1999/02/22-rdf-syntax-ns#>

select ?subclass ?superclass where {
  { ?subclass rdfs:subClassOf ?superclass }
  union
  { ?subclass rdfs:subClassOf [ owl:intersectionOf [ rdf:rest* [ rdf:first ?superclass ] ] ] }
}

--------------------------------------------------------------------------------------
| subclass                                 | superclass                              |
======================================================================================
| <http://www.examples.org/school#Student> | _:b0                                    |
| <http://www.examples.org/school#Student> | <http://www.examples.org/school#Person> |
| <http://www.examples.org/school#Student> | _:b1                                    |
--------------------------------------------------------------------------------------

两个空白节点是匿名交集类,以及匿名限制类(在某些课程中注册).如果您只想要IRI结果,可以使用filter:

prefix rdfs:  <http://www.w3.org/2000/01/rdf-schema#>
prefix owl:   <http://www.w3.org/2002/07/owl#>
prefix rdf:   <http://www.w3.org/1999/02/22-rdf-syntax-ns#>

select ?subclass ?superclass where {
  { ?subclass rdfs:subClassOf ?superclass }
  union
  { ?subclass rdfs:subClassOf [ owl:intersectionOf [ rdf:rest* [ rdf:first ?superclass ] ] ] }

  filter( isIRI( ?superclass ) )
}

--------------------------------------------------------------------------------------
| subclass                                 | superclass                              |
======================================================================================
| <http://www.examples.org/school#Student> | <http://www.examples.org/school#Person> |
--------------------------------------------------------------------------------------

现在,作为最后的润色,如果你想使你的查询有点小,因为在这两个唯一的区别union版图案是连接路径?subclass和?superclass,你其实可以只用一个属性路径写这篇文章.(尽管如Sparql查询Subclass或EquivalentTo中所述,如果你这样做,你可能会遇到Protégé的一些问题.)这个想法是你可以重写这个:

{ ?subclass rdfs:subClassOf ?superclass }
union
{ ?subclass rdfs:subClassOf [ owl:intersectionOf [ rdf:rest* [ rdf:first ?superclass ] ] ] }

因此,通过使用属性路径,这也消除了对空白节点的需求:

?subclass ( rdfs:subClassOf |
            ( rdfs:subClassOf / owl:intersectionOf / rdf:rest* / rdf:first ) ) ?superclass

你可以简化一下

?subclass rdfs:subClassOf/((owl:intersectionOf/rdf:rest*/rdf:first)+) ?superclass

你甚至可以从中删除一个级别的括号来制作它

?subclass rdfs:subClassOf/(owl:intersectionOf/rdf:rest*/rdf:first)+ ?superclass

但是你必须开始记住优先规则,这并不是很有趣.该查询有效:

prefix rdfs:  <http://www.w3.org/2000/01/rdf-schema#>
prefix owl:   <http://www.w3.org/2002/07/owl#>
prefix rdf:   <http://www.w3.org/1999/02/22-rdf-syntax-ns#>

select ?subclass ?superclass where {
  ?subclass rdfs:subClassOf/(owl:intersectionOf/rdf:rest*/rdf:first)+ ?superclass
  filter( isIRI( ?superclass ) )
}

--------------------------------------------------------------------------------------
| subclass                                 | superclass                              |
======================================================================================
| <http://www.examples.org/school#Student> | <http://www.examples.org/school#Person> |
--------------------------------------------------------------------------------------