pythonrdflib

How to get graph's namespaces in rdflib

I've loaded an RDF file in Python, using rdflib:

g = rdflib.Graph()
g.parse ( "foaf_ex.rdf" )

the *.rdf defines a number of prefix/namespace pairs (e.g., foaf:) and I know they come into g because they're still there when I print g.serialize().

My question is, is there an easy way to go get "foaf:Person" resolved from g, i.e., turned into "http://xmlns.com/foaf/0.1/Person"? getting an URIRef straight from the initial prefixed URI would be even better, but it would help anyway if I might get at least the full URI string.

Solution

  • Answering myself to this old question, with the experience gathered meanwhile.

    rdflib has a NamespaceManager class, the Graph object has a namespace_manager property, which can be passed to the a from_n3() function, and the latter does what I need:

    def from_n3(s, default=None, backend=None, nsm=None):
    r'''
    Creates the Identifier corresponding to the given n3 string.

        >>> from_n3('<http://ex.com/foo>') == URIRef('http://ex.com/foo')
        True
        >>> from_n3('"foo"@de') == Literal('foo', lang='de')
        True
        >>> from_n3('"""multi\nline\nstring"""@en') == Literal(
        ...     'multi\nline\nstring', lang='en')
        True
        >>> from_n3('42') == Literal(42)
        True
        >>> from_n3(Literal(42).n3()) == Literal(42)
        True
        >>> from_n3('"42"^^xsd:integer') == Literal(42)
        True
        >>> from rdflib import RDFS
        >>> from_n3('rdfs:label') == RDFS['label']
        True
        >>> nsm = NamespaceManager(Graph())
        >>> nsm.bind('dbpedia', 'http://dbpedia.org/resource/')
        >>> berlin = URIRef('http://dbpedia.org/resource/Berlin')
        >>> from_n3('dbpedia:Berlin', nsm=nsm) == berlin
        True

    '''

    I've developed an extended version of NamespaceManager, the XNamespaceManager, which makes it simple to access this and other functions.