On Feb 15, 10:09 am, Wojciech Muła
<wojciech_m...@poczta.null.onet.pl.invalid> wrote:
> import re
>
> s = 'xxaabbddee'
> m = re.compile("(..)")
> s1 = m.sub("\\1:", s)[:-1]
One can modify this slightly:
s = 'xxaabbddee'
m = re.compile('..')
s1 = ':'.join(m.findall(s))
Depending on one's taste this could be clearer. The more general
answer, from the itertools docs:
from itertools import izip_longest
def grouper(n, iterable, fillvalue=None):
"grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
args = [iter(iterable)] * n
return izip_longest(fillvalue=fillvalue, *args)
s2 = ':'.join(''.join(pair) for pair in grouper(2, s, ''))
Note that this behaves differently to the previous solutions, for
sequences with an odd length.
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