OWL API:如何确保域和范围限制?

时间:2012-04-02 15:50:03

标签: dns range owl restriction

我尝试使用OWL API创建OWL Ontologies。我可以定义它们之间的类,个体和关系。

当我使用域#hasPart和范围#A定义ObjectProperty #B时,我预计此属性只能应用于这两个类的个体。但实际上API并不关心限制,因此可以在类#hasPart的两个成员之间分配#C,例如:

import org.semanticweb.owlapi.apibinding.OWLManager;
import org.semanticweb.owlapi.model.*;

public class OwlTest
{
    public static void main(String[] args)
    throws org.semanticweb.owlapi.model.OWLOntologyStorageException, org.semanticweb.owlapi.model.OWLOntologyCreationException, Exception
    {
        OWLOntologyManager manager = OWLManager.createOWLOntologyManager();
        OWLDataFactory df = manager.getOWLDataFactory();
        OWLOntology o = manager.createOntology();

       //------------------------------------------------------------------

        OWLClass clsA = df.getOWLClass( IRI.create("#A") );
        OWLClass clsB = df.getOWLClass( IRI.create("#B") );
        OWLClass clsC = df.getOWLClass( IRI.create("#C") );

        OWLObjectProperty hasPart = df.getOWLObjectProperty( IRI.create("#hasPart") );
        OWLObjectPropertyDomainAxiom domainAxiom = df.getOWLObjectPropertyDomainAxiom(hasPart, clsA);
        OWLObjectPropertyRangeAxiom   rangeAxiom = df.getOWLObjectPropertyRangeAxiom( hasPart, clsB);

        manager.addAxiom(o, domainAxiom);
        manager.addAxiom(o,  rangeAxiom);

       //------------------------------------------------------------------

        OWLNamedIndividual a1 = df.getOWLNamedIndividual( IRI.create("a1") );
        OWLNamedIndividual b1 = df.getOWLNamedIndividual( IRI.create("b1") );
        OWLNamedIndividual c1 = df.getOWLNamedIndividual( IRI.create("c1") );
        OWLNamedIndividual c2 = df.getOWLNamedIndividual( IRI.create("c2") );

        manager.addAxiom(o, df.getOWLClassAssertionAxiom(clsA, a1));
        manager.addAxiom(o, df.getOWLClassAssertionAxiom(clsB, b1));
        manager.addAxiom(o, df.getOWLClassAssertionAxiom(clsC, c1));
        manager.addAxiom(o, df.getOWLClassAssertionAxiom(clsC, c2));

        manager.addAxiom(o, df.getOWLObjectPropertyAssertionAxiom(hasPart, c1, c2));   // ObjectProperty '#hasPart' should only work for objects from Domain 'clsA' and Range 'clsB'

       //------------------------------------------------------------------

        manager.saveOntology(o, IRI.create("file:/tmp/data.owl"));
    }
}

输出/tmp/data.owl

...
    <ObjectProperty rdf:about="#hasPart">
        <rdfs:domain rdf:resource="#A"/>
        <rdfs:range rdf:resource="#B"/>
    </ObjectProperty>

    <Class rdf:about="#A"/>
    <Class rdf:about="#B"/>
    <Class rdf:about="#C"/>

    <NamedIndividual rdf:about="a1">
        <rdf:type rdf:resource="#A"/>
    </NamedIndividual>

    <NamedIndividual rdf:about="b1">
        <rdf:type rdf:resource="#B"/>
    </NamedIndividual>

    <NamedIndividual rdf:about="c1">
        <rdf:type rdf:resource="#C"/>
        <p1:hasPart rdf:resource="c2"/>
    </NamedIndividual>
...

我现在正在寻找以编程方式处理这种限制的推荐方法..?非常感谢提前!

1 个答案:

答案 0 :(得分:3)

是的,使用hasPart w / C没有任何问题,推理者会假设你最终会告诉它c1也是A,或者C与A相同。

OWL-API不会强制执行您想要的行为,听起来您正在寻找某种完整性约束,就像您进入正常的关系系统一样。您要么必须将其烘焙到您的应用程序中,要么查看Pellet的integrity constraints之类的内容,这些内容将在即将发布的Pellet 3版本中提供,目前可在Stardog中找到。