SPARQLWrapper not returning JSON

Question

The following code should return a dictionary of the subject of all triples in the ontology. It, instead, returns the entire ontology as an XML string.

from SPARQLWrapper import SPARQLWrapper, JSON

sparql = SPARQLWrapper("http://purl.org/sudo/ontology/sudo.owl")
sparql.setQuery("""
    SELECT ?subject
    WHERE {?subject ?verb  ?object}
    """)

sparql.setReturnFormat(JSON)
results = sparql.query().convert()
print results.keys()

The above code works fine with a different ontology, which suggests that the ontology is the issue. I'm not sure what the issue with the ontology could be. I generated the ontology with Protege, it can load into vOWL, and it passes vOWL's ontology validation.

Answer 1

SPARQLWrapper()'s first argument should be a SPARQL endpoint address:

• sparqlwrapper can't perform SPARQL queries against RDF files.
• if you want to query against an RDF file, you should load it into a local store using rdflib.

from rdflib import Graph

g = Graph()
g.parse("http://purl.org/sudo/ontology/sudo.owl", format="xml")

qres = g.query("""
    SELECT DISTINCT ?s {
        ?s ?p ?o
    }""")

for row in qres:
    print("%s" % row)

If you really need SPARQL query results in JSON format (spec):

import sys
from rdflib import Graph
from rdflib.plugins.sparql.results.jsonresults import JSONResultSerializer

g = Graph()
g.parse("http://purl.org/sudo/ontology/sudo.owl", format="xml")

qres = g.query("""
    SELECT DISTINCT ?s {
        ?s ?p ?o
    }""")

JSONResultSerializer(qres).serialize(sys.stdout)

If you wish to abstract from RDF serialization, you should use owlready.