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我在 Eclipse 3.4 中使用 OWL Api 4.0，在 Protege 4 中有一个简单的本体。我有两个类“Ward”和“Gaurdian”。这些类的个体通过对象属性 isWardOf 关联。我如何检索与 Gaurdian 类的同一个人相关的 Ward 类的个人。考虑下图：-

在此处输入图像描述

我想检索 Peter 和 Allice 是亲戚或兄弟姐妹的事实，因为他们都与 Jack 有联系。关于如何使用 OWL API 4.0 实现这一点的任何粗略线索。

我完整的猫头鹰文件已附加：-

<?xml version="1.0"?>


<!DOCTYPE Ontology [
<!ENTITY xsd "http://www.w3.org/2001/XMLSchema#" >
<!ENTITY xml "http://www.w3.org/XML/1998/namespace" >
<!ENTITY rdfs "http://www.w3.org/2000/01/rdf-schema#" >
<!ENTITY rdf "http://www.w3.org/1999/02/22-rdf-syntax-ns#" >
]>


<Ontology xmlns="http://www.w3.org/2002/07/owl#"
 xml:base="http://www.semanticweb.org/antonio/ontologies/2014/11/untitled-ontology-46"
 xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#"
 xmlns:xsd="http://www.w3.org/2001/XMLSchema#"
 xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
 xmlns:xml="http://www.w3.org/XML/1998/namespace"
 ontologyIRI="http://www.semanticweb.org/antonio/ontologies/2014/11/untitled-ontology- 
 46">
 <Prefix name="rdf" IRI="http://www.w3.org/1999/02/22-rdf-syntax-ns#"/>
 <Prefix name="rdfs" IRI="http://www.w3.org/2000/01/rdf-schema#"/>
 <Prefix name="xsd" IRI="http://www.w3.org/2001/XMLSchema#"/>
 <Prefix name="owl" IRI="http://www.w3.org/2002/07/owl#"/>
 <Declaration>
    <Class IRI="#Gaurdian"/>
 </Declaration>
 <Declaration>
    <Class IRI="#Ward"/>
 </Declaration>
 <Declaration>
    <ObjectProperty IRI="#isWardOf"/>
 </Declaration>
 <Declaration>
    <NamedIndividual IRI="#Allice"/>
 </Declaration>
 <Declaration>
    <NamedIndividual IRI="#Amber"/>
 </Declaration>
 <Declaration>
    <NamedIndividual IRI="#Jack"/>
 </Declaration>
 <Declaration>
    <NamedIndividual IRI="#Paul"/>
 </Declaration>
 <Declaration>
     <NamedIndividual IRI="#Peter"/>
 </Declaration>
 <ClassAssertion>
    <Class IRI="#Ward"/>
    <NamedIndividual IRI="#Allice"/>
 </ClassAssertion>
 <ClassAssertion>
    <Class IRI="#Gaurdian"/>
    <NamedIndividual IRI="#Amber"/>
 </ClassAssertion>
 <ClassAssertion>
    <Class IRI="#Gaurdian"/>
    <NamedIndividual IRI="#Jack"/>
 </ClassAssertion>
 <ClassAssertion>
    <Class IRI="#Ward"/>
    <NamedIndividual IRI="#Paul"/>
 </ClassAssertion>
 <ClassAssertion>
    <Class IRI="#Ward"/>
    <NamedIndividual IRI="#Peter"/>
 </ClassAssertion>
 <ObjectPropertyAssertion>
    <ObjectProperty IRI="#isWardOf"/>
    <NamedIndividual IRI="#Allice"/>
    <NamedIndividual IRI="#Jack"/>
 </ObjectPropertyAssertion>
 <ObjectPropertyAssertion>
    <ObjectProperty IRI="#isWardOf"/>
    <NamedIndividual IRI="#Amber"/>
    <NamedIndividual IRI="#Jack"/>
 </ObjectPropertyAssertion>
 <ObjectPropertyAssertion>
    <ObjectProperty IRI="#isWardOf"/>
    <NamedIndividual IRI="#Paul"/>
    <NamedIndividual IRI="#Amber"/>
 </ObjectPropertyAssertion>
 <ObjectPropertyDomain>
    <ObjectProperty IRI="#isWardOf"/>
    <Class IRI="#Ward"/>
  </ObjectPropertyDomain>
  <ObjectPropertyRange>
    <ObjectProperty IRI="#isWardOf"/>
    <Class IRI="#Gaurdian"/>
  </ObjectPropertyRange>
  </Ontology> >
4

2 回答 2

2

这是我能想到的最简单的方法。它涉及用名义进行推理，因此计算成本可能很高。但是，如果本体不是太大，这种方法是可行的。

这个想法是获取每个 Gaudian 的所有实例。然后对于每个这样的个人，通过 isWard 属性获取与其相关的所有个人。如果它们的大小大于 1，那么这些集合就是您要寻找的（如果集合大小为 1，那么给定的 Gaudian 只有一个 Ward）。用于此的 OWL API 代码类似于：

// load an ontology
OWLOntologyManager manager = OWLManager.createOWLOntologyManager();
OWLOntology ontology = manager.loadOntologyFromOntologyDocument(ONTOLOGY_IRI);
OWLDataFactory df = manager.getOWLDataFactory();

// We need a reasoner to ask for individuals
OWLReasoner reasoner = createReasoner(ontology);
reasoner.precomputeInferences(InferenceType.CLASS_ASSERTIONS);

// get all the gaurdians in the ontology
OWLClass gaurdian = df.getOWLClass(IRI.create("#Gaurdian"));
Set<OWLNamedIndividual> gaurdians = reasoner.getInstances(gaurdian, false).getFlattened();


for (OWLNamedIndividual g : gaurdians) {
    // all wards of a given gaurdian g
    OWLObjectProperty isWardOf = df.getOWLObjectProperty(IRI.create("#isWardOf"));
    OWLClassExpression wardsOfG = df.getOWLObjectHasValue(isWardOf, g);
    // get all the wards related to a given gaurdian
    Set<OWLNamedIndividual> wards = reasoner.getInstances(wardsOfG, false).getFlattened();
    if ( wards.size() > 1 ) {
        // this set of wards is connected to the same gaurdian
    }
}
于 2015-01-14T17:47:22.100 回答
0

在 OWL API 文档中，这里有对粗略指南教程源代码的参考

其中一项测试检索对象属性的断言，您应该能够使其适应您的需求：

@Test
public void testIndividualAssertions() throws OWLException {
    OWLOntologyManager m = create();
    OWLOntology o = m.createOntology(EXAMPLE_IRI);
    // We want to state that matthew has a father who is peter.
    OWLIndividual matthew = df.getOWLNamedIndividual(IRI.create(EXAMPLE_IRI
            + "#matthew"));
    OWLIndividual peter = df.getOWLNamedIndividual(IRI.create(EXAMPLE_IRI
            + "#peter"));
    // We need the hasFather property
    OWLObjectProperty hasFather = df.getOWLObjectProperty(IRI
            .create(EXAMPLE_IRI + "#hasFather"));
    // matthew --> hasFather --> peter
    OWLObjectPropertyAssertionAxiom assertion = df
            .getOWLObjectPropertyAssertionAxiom(hasFather, matthew, peter);
    // Finally, add the axiom to our ontology and save
    AddAxiom addAxiomChange = new AddAxiom(o, assertion);
    m.applyChange(addAxiomChange);
    // matthew is an instance of Person
    OWLClass personClass = df.getOWLClass(IRI.create(EXAMPLE_IRI
            + "#Person"));
    OWLClassAssertionAxiom ax = df.getOWLClassAssertionAxiom(personClass,
            matthew);
    // Add this axiom to our ontology - with a convenience method
    m.addAxiom(o, ax);
}
于 2014-12-31T20:34:24.563 回答