Nick Coghlan wrote: > > I think this is about the best you can do for an in-place version: > for i, x in enumerate(reversed(lst)): > if x == 2: > del lst[-i]
Don't think, implement and measure. You may be surprised. Compare these two for example: !def method_del_bkwds(lst, x): ! for inx in xrange(len(lst) - 1, -1, -1): ! if lst[inx] == x: ! del lst[inx] ! return lst !def method_coghlan(lst, x): ! for i, obj in enumerate(reversed(lst)): ! if obj == x: ! del lst[-i] ! return lst > > The effbot's version is still going to be faster though: > lst = [x for x in lst if x != 2] Have you measured this? -- http://mail.python.org/mailman/listinfo/python-list
