我需要一个SPARQL查询来获取所有可用的逆属性。exp(之前,之后,配偶,......等)我在特定域(Person)上试过这个:
SELECT
DISTINCT ?predicate
WHERE
{
?subject a dbo:Person .
?object a dbo:Person .
?subject ?predicate ?object .
?object ?predicate ?subject .
}
结果:
- http://www.w3.org/2000/01/rdf-schema#seeAlso
- http://www.w3.org/2002/07/owl#sameAs
- http://www.w3.org/2002/07/owl#differentFrom
- http://dbpedia.org/property/deathPlace
- http://dbpedia.org/ontology/associatedBand
- http://dbpedia.org/ontology/author
- http://dbpedia.org/ontology/associatedMusicalArtist
- http://dbpedia.org/ontology/field
- http://dbpedia.org/ontology/governor
- http://dbpedia.org/ontology/lastAppearance
- http://dbpedia.org/ontology/managerClub
- http://dbpedia.org/ontology/recordLabel
- http://dbpedia.org/ontology/relation
- http://dbpedia.org/ontology/related
- http://dbpedia.org/ontology/relative
- http://dbpedia.org/ontology/starring
- http://dbpedia.org/property/p
- http://dbpedia.org/property/title
- http://purl.org/linguistics/gold/hypernym