George Sakkis: > Is this a rhetorical question ? If not, try this: It wasn't a rhetorical question.
> >>> x = (i for i in xrange(100) if i&1) > >>> if leniter(x): print x.next() What's your point? Maybe you mean that it consumes the given iterator? I am aware of that, it's written in the function docstring too. But sometimes you don't need the elements of a given iterator, you just need to know how many elements it has. A very simple example: s = "aaabbbbbaabbbbbb" from itertools import groupby print [(h,leniter(g)) for h,g in groupby(s)] Bye, bearophile -- http://mail.python.org/mailman/listinfo/python-list