Sparql skos：更广泛

Question

I'm doing a SPARQL query on the DBpediaset, but I am having some issues (due to lack of detailed SPARQL knowledge) with a query limitation:

I first 'get' all music artists:

?person rdf:type <http://dbpedia.org/ontology/MusicalArtist> .

But I want to limit this to the broader category Category:American_musicians (via traversing skos:broader ?): how?

*= while the question is specific, I've encountered this quest many times when wanting to running sparql queries.

Answer 1

This can be made easier with property paths in SPARQL 1.1

SELECT DISTINCT ( ?person )
WHERE
{
  ?person rdf:type dbpedia-owl:MusicalArtist .
  ?person skos:subject  skos:broader* category:American_musicians  .
}

Here it displays all the ancestors that could be reached via the skos:broader property.

Answer 2

I'm amazed this simple question hasn't been answered correctly in 3 years, and how much uncertainty and doubt people spread.

SELECT * { ?person a dbo:MusicalArtist . filter exists {?person dct:subject/skos:broader* dbc:American_musicians} }

• corrected a few prefixes: dbo instead of the long dbpedia-owl , dbc instead of category . These short prefixes are builtin to DBpedia
• corrected skos:subject (no such prop exists) to dct:subject
• corrected the query with property paths, it was missing /
• skos:broader is not transitive, skos:broaderTransitive is. However, DBpedia doesn't have the latter (no transitive reasoning)
• replaced DISTINCT which is expensive with FILTER EXISTS which is much faster. The FILTER can stop at the first relevant sub-category it finds, while the original query first finds all such sub-cats per artist, then discards them ( DISTINCT ), sorts the artists in memory and removes duplicates.

Answer 3

There's no really good way to do this, but here's a verbose way:

SELECT DISTINCT ( ?person )
WHERE
{
  ?person rdf:type dbpedia-owl:MusicalArtist .
  {
    ?person skos:subject [ skos:broader category:American_musicians ] .
  } UNION {
    ?person skos:subject [ skos:broader [ skos:broader category:American_musicians ] ] .
  } UNION {
    ?person skos:subject [ skos:broader [ skos:broader [ skos:broader category:American_musicians ] ] ] .
  } UNION {
    ?person skos:subject [ skos:broader [ skos:broader [ skos:broader [ skos:broader category:American_musicians ] ] ] ] .
  } UNION {
    ?person skos:subject [ skos:broader [ skos:broader [ skos:broader [ skos:broader [ skos:broader category:American_musicians ] ] ] ] ] .
  } UNION {
    ?person skos:subject [ skos:broader [ skos:broader [ skos:broader [ skos:broader [ skos:broader [ skos:broader category:American_musicians ] ] ] ] ] ] .
  } UNION {
    ?person skos:subject [ skos:broader [ skos:broader [ skos:broader [ skos:broader [ skos:broader [ skos:broader [ skos:broader category:American_musicians ] ] ] ] ] ] ] .
  }
}

For figuring out how many levels you need, you can change SELECT DISTINCT to SELECT COUNT DISTINCT and stop adding levels when the count stops going up.

Answer 4

This is really easy to perform in neo4j. An alternative to accomplish your task in SPARQL could be to extract all the subgraph under "Category:American_musicians" by iterating via code on subcategories.

Eg. pseudo code in java would be something like:

String startCategory = "<http://dbpedia.org/resource/Category:American_musicians>";
iterateTraversalFunction(startCategory);

then the traversal function would be:

public void iterateTraversalFunction(String startCategory){
     ArrayList<String> artistsURI = // SPARQL query ?person skos:subject startCategory . ?person rdf:type MusicalArtist 

    ArrayList<String> subCategoriesURI = // SPARQL query ?subCat skos startCategory
    // Repeat recursively
   for(String subCatURI: subCategoriesURI){
       iterateTraversalFunction(subCatURI);
   }
}

Hope this helps, - Dan