Java示例

Question

我有一个语义网络。是否可以使用jena框架基于某些规则在语义Web中创建新对象。例如，有一个对象具有某个属性，那么您需要创建一个新对象并在它们之间建立连接。有可能吗？

Answer 1

是的，这在耶拿的规则系统中是可行的。通常，我们使用makeSkolem Reasoner Builtin Primitive：

创建此类节点

[example:
  (?a urn:ex:owns ?b) 
  makeSkolem(?ownership,?a,?b)
  ->
  (?a urn:ex:hasOwnership ?ownership)
  (?ownership urn:ex:of ?b)
]

这将在图表中创建一个新的空白节点，用于统一<urn:ex:owns>三元组。例如，当给出包含三<urn:ex:a> <urn:ex:owns> <urn:ex:b>个输入的图形时，前面的规则将生成以下图形结构：

<urn:ex:a> <urn:ex:hasOwnership> [
  <urn:ex:of> <urn:ex:b>
].

如果你有一些生成它们的方案，你也可以在你的规则中构建URI。

Java示例

假设您的类路径中存在so.rules并包含上面的规则，以下Java代码将演示此任务的自定义规则。

// Obtains a list of rules to pass to a rule-based reasoner
// These rules are read from a file.
// This is the most common case.
final List<Rule> rules;
try (final InputStream src = Resources.getResource("so.rules").openStream()) {
  rules = Rule
      .parseRules(Rule.rulesParserFromReader(new BufferedReader(new InputStreamReader(src))));
}

// Create a rule-based reasoner.
// There are multiple types of reasoners available.
// You may prefer some over others (e.g., when performing OWL inference in tandem with custom rules)
final GenericRuleReasoner reasoner =
    (GenericRuleReasoner) GenericRuleReasonerFactory.theInstance().create(null);
reasoner.setRules(rules);

// Create a RDF Model to store data in.
// Create an inference model to interact with.
// The inference model will store any added data in the base model.
// The inference model will store inferred triples internally.
final Model baseModel = ModelFactory.createDefaultModel();
final InfModel model = ModelFactory.createInfModel(reasoner, baseModel);
model.prepare();

// Stimulate the rule by introducing the desired triples to the graph
// :a :owns :b
final Property owns = model.createProperty("urn:ex:", "owns");
final Property hasOwnership = model.createProperty("urn:ex:","hasOwnership");
final Property of = model.createProperty("urn:ex:","of");

final Resource a = model.createResource("urn:ex:a");
final Resource b = model.createResource("urn:ex:b");
model.add(a,owns,b);

// Verify that the rule has fired. That is, that we have created some node
// and that the node relates our other two resources
// -> :a :hasOwnership [ :of :b ]
assertTrue(a.hasProperty(hasOwnership));
final Resource createdObject = a.getPropertyResourceValue(hasOwnership);
assertTrue(createdObject.hasProperty(of,b));

Answer 2

如果您需要相当简单，可以使用SPARQL CONSTRUCT查询，即

CONSTRUCT { ?p :hasGrandfather ?g . }

WHERE {
   ?p      :hasParent ?parent .
   ?parent :hasParent ?g .
   ?g      :gender    :male .
}

会因为陈述祖父关系而产生三元组。

如果您的需求更加复杂，则可以使用SHACL来实现此目标，其中Jena顶部存在implementation。我将举一个简短的例子。假设您有以下RDF数据：

@prefix ex: <http://example.com/ns#> .
@prefix rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#> .
@prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#> .

ex:InvalidRectangle
    a ex:Rectangle .

ex:NonSquareRectangle
    a ex:Rectangle ;
    ex:height 2 ;
    ex:width 3 .

ex:SquareRectangle
    a ex:Rectangle ;
    ex:height 4 ;
    ex:width 4 .

您为其定义以下形状文件：

@prefix ex: <http://example.com/ns#> .
@prefix rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#> .
@prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#> .
@prefix dash: <http://datashapes.org/dash#> .
@prefix sh: <http://www.w3.org/ns/shacl#> .
@prefix xsd: <http://www.w3.org/2001/XMLSchema#> .

ex:Rectangle
    a rdfs:Class, sh:NodeShape ;
    rdfs:label "Rectangle" ;
    sh:property [
        sh:path ex:height ;
        sh:datatype xsd:integer ;
        sh:maxCount 1 ;
        sh:minCount 1 ;
        sh:name "height" ;
    ] ;
    sh:property [
        sh:path ex:width ;
        sh:datatype xsd:integer ;
        sh:maxCount 1 ;
        sh:minCount 1 ;
        sh:name "width" ;
    ] ;
    sh:rule [
        a sh:TripleRule ;
        sh:subject sh:this ;
        sh:predicate rdf:type ;
        sh:object ex:Square ;
        sh:condition ex:Rectangle ;
        sh:condition [
            sh:property [
                sh:path ex:width ;
                sh:equals ex:height ;
            ] ;
        ] ;
    ] .

它将生成以下RDF数据：

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

<http://example.com/ns#SquareRectangle>
        a       <http://example.com/ns#Square> .

您可以将其添加到RDF商店。

此示例包含代码here，以及更多advanced example。

使用推理规则创建新对象

2 个答案:

Java示例