RetroSearch Browse

Wed Aug 31 07:30:58 CEST 2005 · https://mail.python.org/pipermail/python-dev/2005-August/055893.html

Steve Holden <steve at holdenweb.com> wrote:
> 
> Guido van Rossum wrote:
> > On 8/30/05, Andrew Durdin <adurdin at gmail.com> wrote:
> [confusion]
> > 
> > 
> > Hm. The example is poorly chosen because it's an end case. The
> > invariant for both is (I'd hope!)
> > 
> >   "".join(s.partition()) == s == "".join(s.rpartition())
> > 
> > Thus,
> > 
> >   "a/b/c".partition("/") returns ("a", "/", "b/c")
> > 
> >   "a/b/c".rpartition("/") returns ("a/b", "/", "c")
> > 
> > That can't be confusing can it?
> > 
> > (Just think of it as rpartition() stopping at the last occurrence,
> > rather than searching from the right. :-)
> > 
> So we can check that a substring x appears precisely once in the string 
> s using
> 
> s.partition(x) == s.rpartition(x)
> 
> Oops, it fails if s == "". I can usually find some way to go wrong ...

There was an example in the standard library that used "s.find(y) ==
s.rfind(y)" as a test for zero or 1 instances of the searched for item.

Generally though, s.count(x)==1 is a better test.

 - Josiah

Home - News ( United States | United Kingdom | Italy | Germany ) - Football scores

Showing content from https://mail.python.org/pipermail/python-dev/2005-August/055893.html below:

[Python-Dev] Remove str.find in 3.0?