On Mon, Oct 11, 2010 at 05:35:21AM -0700, Ethan Furman wrote: > Antoon Pardon wrote: > >On Sat, Oct 09, 2010 at 01:37:03AM +0000, Steven D'Aprano wrote: > > > >>On Fri, 08 Oct 2010 15:53:17 -0400, Jed Smith wrote: > >> > > > >I stand by that claim. I think it was fairly obvious that what I meant > >was that it was impossible to give such a numeric value that would work > >with arbitrary L. > > > >if L2 == list(reversed(L1)) and a and b are in the range 1 < x <= len(L), > >we have the following invariant. > > > > L1[a:b] == L2[b-1:a-1:-1] > > Are you sure?
I'm sorry, I was not careful enough in writing this down. It should be that the invariant holds for 1 <= x < len >>> a=3 >>> b=7 >>> L1=['e', 'g', 'h', 'k', 'o', 'p', 'r', 't', 'x', 'z'] >>> L2 = list(reversed(L1)) >>> L1 ['e', 'g', 'h', 'k', 'o', 'p', 'r', 't', 'x', 'z'] >>> L2 ['z', 'x', 't', 'r', 'p', 'o', 'k', 'h', 'g', 'e'] >>> L1[a:b] ['k', 'o', 'p', 'r'] >>> L2[b-1:a-1:-1] ['k', 'o', 'p', 'r'] > >However this no longer works if either nr is 0. > > In which case it's not an invariant, is it? The point being that the breaking of the invariant at that point is IMO a sign of bad design. It makes it difficult to use the reversed slice in a general way. More specifically, if you have a function that produces an a1 and b1, that in combination with a list, always gives you the desired slice by writing L[a1:b1]. There is no straigtforward way to transform this a1 and b1 into a2 and b2, so that L[a2:b2:-1] is the reversed of L[a1:b1] -- Antoon Pardon -- http://mail.python.org/mailman/listinfo/python-list
