15-08-2009 candide <cand...@free.invalid> wrote:
Suppose you need to split a string into substrings of a given size
(except
possibly the last substring). I make the hypothesis the first slice is
at the end of the string.
A typical example is provided by formatting a decimal string with
thousands separator.
I'd use iterators, especially for longer strings...
import itertools
def separate(text, grouplen=3, sep=','):
"separate('12345678') -> '123,456,78'"
repeated_iterator = [iter(text)] * grouplen
groups = itertools.izip_longest(fillvalue='', *repeated_iterator)
strings = (''.join(group) for group in groups) # gen. expr.
return sep.join(strings)
def back_separate(text, grouplen=3, sep=','):
"back_separate('12345678') -> '12,345,678'"
repeated_iterator = [reversed(text)] * grouplen
groups = itertools.izip_longest(fillvalue='', *repeated_iterator)
strings = [''.join(reversed(group)) for group in groups] # list compr.
return sep.join(reversed(strings))
print separate('12345678')
print back_separate('12345678')
# alternate implementation
# (without "materializing" 'strings' as a list in back_separate):
def separate(text, grouplen=3, sep=','):
"separate('12345678') -> '12,345,678'"
textlen = len(text)
end = textlen - (textlen % grouplen)
repeated_iterator = [iter(itertools.islice(text, 0, end))] * grouplen
strings = itertools.imap(lambda *chars: ''.join(chars),
*repeated_iterator)
return sep.join(itertools.chain(strings, (text[end:],)))
def back_separate(text, grouplen=3, sep=','):
"back_separate('12345678') -> '12,345,678'"
beg = len(text) % grouplen
repeated_iterator = [iter(itertools.islice(text, beg, None))] *
grouplen
strings = itertools.imap(lambda *chars: ''.join(chars),
*repeated_iterator)
return sep.join(itertools.chain((text[:beg],), strings))
print separate('12345678')
print back_separate('12345678')
http://docs.python.org/library/itertools.html#recipes
was the inspiration for me (especially grouper).
Cheers,
*j
--
Jan Kaliszewski (zuo) <z...@chopin.edu.pl>
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