Slicing [N::-1]
Mensanator
mensanator at aol.com
Sat Mar 6 03:09:59 CET 2010
On Mar 5, 6:34 pm, Gary Herron <gher... at islandtraining.com> wrote:
> Mensanator wrote:
> > On Mar 5, 3:42 pm, Gary Herron <gher... at islandtraining.com> wrote:
>
> >> Mensanator wrote:
>
> >>> The only way to get a 0 from a reverse range() is to have a bound of
> >>> -1.
>
> >> Not quite. An empty second bound goes all the way to the zero index:
>
> > Not the same thing. You're using the bounds of the slice index.
> > I was refering to the bounds of the range() function.
>
> >>>> for a in range(9,-9,-1):print(a,end=' ')
>
> > 9 8 7 6 5 4 3 2 1 0 -1 -2 -3 -4 -5 -6 -7 -8
>
> > To get that to stop at 0, you use a -1 as the bounds:
>
> >>>> for a in range(9,-1,-1):print(a,end=' ')
>
> > 9 8 7 6 5 4 3 2 1 0
>
> > Your slice notation only works if the last (first?) number in
> > the range happens to be 0. What if the range bounds were variables?
> > You may still want to force the range's last number to be 0 by
> > using a constant like range(a,-1,-1) rather than just take
> > the last number of range(a,b,-1) by using slice notation.
>
> All true and valid of course, but I was just contridicting the "the
> ONLY way to get a 0" (emphasis mine) part of the statement.
Does it still contradict if you do not use the '::' as the OP
requested?
>
> Gary Herron
>
>
>
>
>
> >> >>> range(9)[2::-1]
> >> [2, 1, 0]
>
> >> Gary Herron- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
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