lbolla <lbo...@gmail.com> writes:
> for k, g in groupby(clean_up(data) , key=lambda s: s.startswith('VLAN')):
> if k:
> key = list(g)[0].replace('VLAN','')This is the nicest solution, I think. Mine was more cumbersome. -- http://mail.python.org/mailman/listinfo/python-list
