Pythonic way to determine if one char of many in a string

rdmurray at bitdance.com rdmurray at bitdance.com
Sat Feb 21 23:24:51 CET 2009


odeits <odeits at gmail.com> wrote:
> On Feb 21, 12:47=A0am, "Gabriel Genellina" <gagsl-... at yahoo.com.ar>
> wrote:
> > En Sat, 21 Feb 2009 01:14:02 -0200, odeits <ode... at gmail.com> escribi=F3:
> >
> > > On Feb 15, 11:31=A0pm, odeits <ode... at gmail.com> wrote:
> > >> It seems what you are actually testing for is if the intersection of
> > >> the two sets is not empty where the first set is the characters in
> > >> your word and the second set is the characters in your defined string.
> >
> > > To expand on what I was saying I thought i should provide a code
> > > snippet:
> >
> > > WORD = 'g' * 100
> > > WORD2 = 'g' * 50 + 'U'
> > > VOWELS = 'aeiouAEIOU'
> > > BIGWORD = 'g' * 10000 + 'U'
> >
> > > def set_test(vowels, word):
> >
> > >  vowels = set( iter(vowels))
> > >  letters = set( iter(word) )
> >
> > >  if letters & vowels:
> > >      return True
> > >  else:
> > >     return False
> >
> > > with python 2.5 I got 1.30 usec/pass against the BIGWORD
> >
> > You could make it slightly faster by removing the iter() call:
> > letters = set(word)
> > And (if vowels are really constant) you could pre-build the vowels set.
> 
> set(word) = set{[word]} meaning a set with one element, the string
> the call to iter makes it set of the letters making up the word.

Did you try it?

Python 2.6.1 (r261:67515, Jan  7 2009, 17:09:13) 
[GCC 4.3.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> set('abcd')
set(['a', 'c', 'b', 'd'])

--RDM




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