> itertools only looks for changes to the key value (the one returned by > operator.itemgetter(0) in your case); it doesn't sort the list for you. > > this should work: > > for k, g in itertools.groupby(sorted(vals), operator.itemgetter(0)): > print k, [i for i in g]
footnote: to turn the contents in an iterator into a list object, list(g) is a bit more convenient than [i for i in g]. </F> -- http://mail.python.org/mailman/listinfo/python-list
