Finding and splitting by base URI

[d.c.so...@gmail.com]

Hello,

I'd like to search a large data set for all terms of a base URI. Can this be done with rdflib?

Something like

for s,p,o in g.triples( (None, DEFAULT['None'], None) ):

    print("{} {} {}".format(s, p, o))

?

And then to split the term, Instead of

<http://some.default.base/#InterestingTerm>

just

InterestingTerm

[Nicholas Car]

Hi D.C,

Can I just reflect back the question at you to see if it is understood? Are you asking for something like this:

import rdflib

# 1
base_uri = 'http://www.w3.org/ns/dx/prof/‘

# 2
g = rdflib.Graph().parse('prof.ttl', format='turtle’) 

# 3
for s, p, o in g.triples((None, None, None)):

    # 4
    if str(s).startswith(base_uri):
        print('s {} - {}'.format(str(s).split('/')[-1], s))
    elif str(p).startswith(base_uri):
        print('p {} - {}'.format(str(p).split('/')[-1], p))
    elif str(o).startswith(base_uri):
        print('o {} - {}'.format(str(o).split('/')[-1], o))

So here, following numbered comments in the code:

1. Identifying a base URI

  - which is the base URI of the W3C’s Profiles Vocabulary, see https://www.w3.org/TR/dx-prof/

2. Reading an RDF source 

  - the file prof.ttl which contains the Profiles Vocabulary

3. Looping through all triples in theta source

  - which is the Profiles Vocabulary in a file, but could be a huge dataset

4. Testing to see if each part (here a Subject s) starts with a base_uri

  - splits and grits it if it does

Are there other things you want to be doing or is this it?

Cheers,

Nick

-- 
http://github.com/RDFLib
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[Boris Pelakh]

If you are just looking to strip the leading namespace from any URI string, you could do the following:

import re

from rdflib import URIRef

def simplify(term):

  return re.sub(r'^.*[#/]', '', str(term)) if isinstance(term, URIRef) else str(term)

for s,p,o in g.triples( (None, DEFAULT['None'], None) ):

    print("{} {} {}".format(simplify(s), simplify(p), simplify(o)))