On 2016-03-24, BartC <b...@freeuk.com> wrote: > On 24/03/2016 15:03, Jon Ribbens wrote: >> On 2016-03-24, BartC <b...@freeuk.com> wrote: >>> On 24/03/2016 14:08, Jon Ribbens wrote: >>>> if you thought you needed to do that then most likely what you >>>> should actually be doing is re-writing your code so you no longer >>>> need to. However, you could do: >>>> >>>> L[:] = [0] * len(L) >>> >>> OK, but that's just building a new list as I've already mentioned. >> >> No it isn't, it's replacing the elements in-place, > > Replace them with what, if not an entirely new list built from > '[0]*len(L)'? (And which would briefly require twice the memory occupied > by the old list, if I'm not mistaken.)
It's replacing them with elements from an entirely new list, which is then discarded, and only the original list remains. -- https://mail.python.org/mailman/listinfo/python-list
