On May 8, 11:14 pm, Ned Deily <n...@acm.org> wrote: > In article <7xprejoswg....@ruckus.brouhaha.com>, > Paul Rubin <http://phr...@nospam.invalid> wrote: > > > > > > > Ross <ross.j...@gmail.com> writes: > > > I have a really long list that I would like segmented into smaller > > > lists. Let's say I had a list a = [1,2,3,4,5,6,7,8,9,10,11,12] and I > > > wanted to split it into groups of 2 or groups of 3 or 4, etc. Is there > > > a way to do this without explicitly defining new lists? > > > That question comes up so often it should probably be a standard > > library function. > > > Anyway, here is an iterator, if that's what you want: > > >>> from itertools import islice > > >>> a = range(12) > > >>> xs = iter(lambda x=iter(a): list(islice(x,3)), []) > > >>> print list(xs) > > [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10, 11]] > > Of course, as the saying goes, there's more than one way to do it ;-) > > python2.6 itertools introduces the izip_longest function and the grouper > recipe <http://docs.python.org/library/itertools.html>: > > def grouper(n, iterable, fillvalue=None): > "grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx" > args = [iter(iterable)] * n > return izip_longest(fillvalue=fillvalue, *args) > > -- > Ned Deily, > n...@acm.org- Hide quoted text - > > - Show quoted text -
Here's a version that works pre-2.6: >>> grouper = lambda iterable,size,fill=None : >>> zip(*[(iterable+[fill,]*(size-1))[i::size] for i in range(size)]) >>> a = range(12) >>> grouper(a,6) [(0, 1, 2, 3, 4, 5), (6, 7, 8, 9, 10, 11)] >>> grouper(a,5) [(0, 1, 2, 3, 4), (5, 6, 7, 8, 9), (10, 11, None, None, None)] >>> grouper(a,3) [(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11)] -- Paul -- http://mail.python.org/mailman/listinfo/python-list
