On 23 сен, 12:48, Peter Otten <__pete...@web.de> wrote: > blumenkraft wrote: > > I have some list: > > x = [8, 9, 1, 7] > > and list of indices I want to delete from x: > > indices_to_delete = [0, 3], so after deletion x must be equal to [9, > > 1]. > > > What is the fastest way to do this? Is there any builtin? > > Why's that obsession with speed? > > >>> items = ["a", "b", "c", "d"] > >>> delenda = [0, 3] > >>> for i in sorted(delenda, reverse=True): > > ... del items[i] > ...>>> items > > ['b', 'c'] > > >>> items = ["a", "b", "c", "d"] > >>> delenda = set([0, 3]) > >>> items = [item for index, item in enumerate(items) if index not in > delenda] > >>> items > > ['b', 'c'] > > If you really need to modify the list in place change > > items = [item for ...] > > to > > items[:] = [item for ...] > > Try these and come back to complain if any of the above slows down your > script significantly... > > Peter
Thanks, it helped (I didn't know about keyword argument "reverse" in sorted function) -- http://mail.python.org/mailman/listinfo/python-list
