我想用一个简短的序号来指定一个地段,但我有麻烦。 我认为,答案必须非常简单。 这里是犯罪法典:
g = rdflib.parse("some_rdf.rdf")
rdf=rdflib.Namespace("http://www.w3.org/1999/02/22-rdf-syntax-ns#")
print "Name Spaces:"
for ns in g.namespaces():
print ns
print "Matching Triples"
print "length of type full uri",len([i for i in g.triples((None,rdflib.term.URIRef( http://www.w3.org/1999/02/22-rdf-syntax-ns#type ),None))])
print "length of type truncated uri",len([i for i in g.triples((None,rdflib.term.URIRef( rdf:type ),None))])
print "length of type , using namespace",len([i for i in g.triples((None,rdf.type,None))])
And the output is:
Name Spaces:
( xml , rdflib.term.URIRef( http://www.w3.org/XML/1998/namespace ))
(u foaf , rdflib.term.URIRef( http://xmlns.com/foaf/0.1/ ))
(u z , rdflib.term.URIRef( http://www.zotero.org/namespaces/export# ))
( rdfs , rdflib.term.URIRef( http://www.w3.org/2000/01/rdf-schema# ))
(u bib , rdflib.term.URIRef( http://purl.org/net/biblio# ))
(u dc , rdflib.term.URIRef( http://purl.org/dc/elements/1.1/ ))
(u prism , rdflib.term.URIRef( http://prismstandard.org/namespaces/1.2/basic/ ))
( rdf , rdflib.term.URIRef( http://www.w3.org/1999/02/22-rdf-syntax-ns# ))
(u dcterms , rdflib.term.URIRef( http://purl.org/dc/terms/ ))
Matching Triples
length of type full uri 132
length of type truncated uri 0 !!!This is wrong should be 132
length of type , using namespace 132
我做了什么错误?