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:
On Fri, Oct 8, 2010 at 1:26 PM, Steven D'Aprano
<st...@remove-this-cybersource.com.au> wrote:
On Fri, 08 Oct 2010 10:21:16 +0200, Antoon Pardon wrote:
Personnaly I find it horrible
that in the following expression: L[a:b:-1], it is impossible to give
a numeric value to b, that will include L[0] into the reversed slice.
L = [1, 2, 3, 4, 5]
L[5:-6:-1]
[5, 4, 3, 2, 1]
a = [1, 2, 3, 4, 5, 6]
a[::-1]
[6, 5, 4, 3, 2, 1]
Well of course that works, that is the standard Python idiom for
reversing a sequence.
But the point was that Antoon claimed that there is no numeric value for
the end position that will include L[0] in the reversed slice. My example
shows that this is not correct.
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?
Python 2.5.4 (r254:67916, Dec 23 2008, 15:10:54) [MSC v.1310 32 bit
(Intel)] on win32
Type "help", "copyright", "credits" or "license" for more information.
--> L1 = [1, 2, 3, 4, 5]
--> L2 = list(reversed(L1))
--> L1
[1, 2, 3, 4, 5]
--> L2
[5, 4, 3, 2, 1]
--> L1[2:5:1], L2[4:1:-1]
([3, 4, 5], [1, 2, 3])
However this no longer works if either nr is 0.
In which case it's not an invariant, is it?
~Ethan~
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