我有这个rdf文件:
<!DOCTYPE rdf:RDF [
<!ENTITY db "http://dbpedia.org/ontology/" >
<!ENTITY owl "http://www.w3.org/2002/07/owl#" >
<!ENTITY xsd "http://www.w3.org/2001/XMLSchema#" >
<!ENTITY rdfs "http://www.w3.org/2000/01/rdf-schema#" >
<!ENTITY rdf "http://www.w3.org/1999/02/22-rdf-syntax-ns#" >]>
<rdf:RDF xmlns="http://dbpedia.org/ontology/"
xml:base="http://dbpedia.org/ontology/"
xmlns:db="http://dbpedia.org/ontology/"
xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#"
xmlns:owl="http://www.w3.org/2002/07/owl#"
xmlns:xsd="http://www.w3.org/2001/XMLSchema#"
xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#">
<owl:ObjectProperty rdf:about="&db;actingHeadteacher">
<rdfs:label xml:lang="el">διευθυντής σχολείου</rdfs:label>
<rdfs:label xml:lang="en">acting headteacher</rdfs:label>
</owl:ObjectProperty>
</rdf:RDF>
并希望按其lang值过滤Literal对象。例如:
from rdflib import Graph
from rdflib.namespace import RDFS
filetype = util.guess_format(rdf_file)
g = Graph()
g.parse(rdf_file, format = filetype)
for s,p,o in g.triples((None, RDFS.label, None)):
print(repr(o)) # rdflib.term.Literal('acting headteacher', lang='en')
# rdflib.term.Literal('διευθυντής σχολείου', lang='el')
我想仅在lang =&#39; en&#39;
的情况下查询o答案 0 :(得分:4)
当您检查manual for rdflib时,您会发现rdflib.term.Literal
有一个名为language
的属性以及一种方法。但是,调用该方法似乎并不适合我。
这样的事情可以做到:
# from rdflib import URIRef
subject = URIRef('&db;actingHeadteacher')
# just getting your literals directly here:
generator = graph.objects(subject, RDFS.label)
for lit in generator:
print lit.language
仅label
或preferredLabel
如果您只对标签/首选标签(SKOS或RDFS)感兴趣,请检查page 47 from the manual:
subject = URIRef('&db;actingHeadteacher')
graph.preferredLabel(subject=subject, label='en') # or label='el'
这会返回(labelProp, label)
对的列表,其中labelProp
为skos:prefLabel
或rdfs:label
。
答案 1 :(得分:0)
在rdflib中可能有更优雅/高性能的解决方案,但您可以使用SPARQL查询:
g = Graph()
g.parse("../stw.nt", format="nt")
qres = g.query(
"""SELECT ?label
WHERE {
?s ?p ?label
FILTER langMatches( lang(?label), "en" )
}"""
)
for row in qres:
print(row.label)