Make string operations on URIRefs return URIRefs

[dgerber]

Currently, concatenation, formatting or %-substitution demote URIRefs to unicode. It would be nice if they behaved like this instead:

In [13]: u = URIRef('/books/{book}/chapters/{chapter}')

In [14]: u.format(book='123', chapter=3)
Out[14]: rdflib.term.URIRef(u'/books/123/chapters/3')

In [15]: 'http://example.org' + _ + '#frag'
Out[15]: rdflib.term.URIRef(u'http://example.org/books/123/chapters/3#frag')

In [16]: URIRef('/books/%s/chapters/%s') % (23, 45)
Out[16]: rdflib.term.URIRef(u'/books/23/chapters/45')

Implementation:

class URIRef(Identifier):

def __add__(self, other):
    return self.__class__(unicode(self) + other)

def __radd__(self, other):
    return self.__class__(other + unicode(self))

def format(self, *args, **kwargs):
    return self.__class__(unicode(self).format(*args, **kwargs))

def __mod__(self, other):
    return self.__class__(unicode(self) % other)

This __mod__ conflicts with property paths unary operators *, ? and + in rdflib_sparql. These could be expressed with (*, __mul__, same precedence as %) or (**, __pow__, higher precedence) instead.

[dgerber]

Am I missing any disadvantage of using * for property path quantifiers, and assigning to % the same meaning as it has in the unicode superclass?

[gromgull]

dgerber: I am very sorry I have no idea why I closed this - I tried to close something else :)

I'll make the changes soon!

[gromgull]

Now where is the "reopen pull request button" ...

[gromgull]

Merged! Thanks for the contribution - really sorry that I just closed without a word :)

[uholzer]

Wait! Doesn't this conflict with #266? Most template strings for format are invalid URIs, since they contain characters like <, > and {, }. As a consequence, the example from dgerber fails:

>>> import rdflib
>>> rdflib.URIRef('/books/{book}/chapters/{chapter}')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/home/urs/.local/lib/python3.2/site-packages/rdflib/term.py", line 198, in __new__
    raise Exception('%s does not look like a valid URI, perhaps you want to urlencode it?')
Exception: %s does not look like a valid URI, perhaps you want to urlencode it?

I think it is better to not implement format on URIRefs and instead use

f = '/books/{book}/chapters/{chapter}'
URIRef(f.format('a','b'))

[gromgull]

Hmm - good point - I only tested it with "%d" - which is fine.

Shame, since I quite like generating URIs by formatting.

Maybe a better solution is to add this to the namespace class - for some reason I do not understand namespace subclasses URIRef, this causes lots of problems. If you remove the subclassing and add formatting to namespace?

>>> Namespace("http://example.org/persons/{id}/resource") % 32

??

[dgerber]

%-substitution won't play well either with %-encoding...

So if URIRef is to be more restrictive, I agree that the Namespace solution (or some URIPattern class) is better.

The only feature I'm actually using is + to build hierarchical uris:

>>> b = URIRef('http://example.org/books/1984')
>>> chapter = b + '/chapters/1'

[gromgull]

How about this?

I removed the format method from URIRef, and made a new URIpattern class (see 249912e)