Alex wrote: > import collections > def tally(seq): > d = collections.defaultdict(int) > for item in seq: > d[item] += 1 > return dict(d)
I'll stop lurking and submit the following: def tally(seq): return dict((group[0], len(tuple(group[1]))) for group in itertools.groupby(sorted(seq))) In general itertools.groupby() seems like a very clean way to do this sort of thing, whether you want to end up with a dict or not. I'll go so far as to suggest that the existence of groupby() obviates the proposed tally(). Maybe I've just coded too much SQL and it has warped my brain... OTOH the latter definition of tally won't cope with iterables, and it seems like O(nlogn) rather than O(n). cheers, Jess _______________________________________________ Python-Dev mailing list Python-Dev@python.org http://mail.python.org/mailman/listinfo/python-dev Unsubscribe: http://mail.python.org/mailman/options/python-dev/archive%40mail-archive.com