On Oct 31, 7:32 pm, [EMAIL PROTECTED] wrote:
> Hrvoje Niksic:
>
> > I'm surprised that no one has proposed a regex solution, such as:
>
> I presume many Python programmers aren't much used in using REs.
>
> > >>> import re
> > >>> re.sub(r'\d{1,3}(?=(?:\d{3})+$)', r'\g<0>.', str(1234567))
> > '1.234.567'
>
> It works with negative numbers too. It's a very nice solution of a
> simple problem that shows what you can do with REs. But I think that
> RE has to be expanded & splitted (and maybe even commented) with a
> VERBOSE, to improve readability, maybe somethign like:
>
> \d {1,3}
> (?= # lockahead assertion
> (?: \d {3} )+ $ # non-group
> )
>
> Bye,
> bearophile
That's 3 times faster on my box and works for negatives too:
def localize(num, sep='.'):
d,m = divmod(abs(num),1000)
return '-'*(num<0) + (localize(d)+sep+'%03d'%m if d else str(m))
George
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