[EMAIL PROTECTED] wrote: > If you want something less easy to understand, you can try fixing the > following code that doesn't work: > > from itertools import groupby > l = [1,2,3,5,6,8,12] > keyf = lambda (pos,x): x-l[pos]>1 > [[el[1] for el in gr] for he,gr in groupby(enumerate(l[:-1]), keyf)]
>>> from itertools import groupby >>> items = [1, 2, 3, 5, 6, 8, 12] >>> [[v for i, v in group] for key, group in groupby(enumerate(items), lambda (i, v): i-v)] [[1, 2, 3], [5, 6], [8], [12]] which is almost identical to the last example in http://docs.python.org/lib/itertools-example.html Peter -- http://mail.python.org/mailman/listinfo/python-list
