Slicing [N::-1]
in [Python]
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From: Mensanator on 5 Mar 2010 19:09 On Mar 5, 3:42 pm, Gary Herron <gher...(a)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. > > >>> range(9)[2::-1] > [2, 1, 0] > > Gary Herron
From: Gary Herron on 5 Mar 2010 19:34 Mensanator wrote: > On Mar 5, 3:42 pm, Gary Herron <gher...(a)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. Gary Herron > >> >>> range(9)[2::-1] >> [2, 1, 0] >> >> Gary Herron >>
From: Mensanator on 5 Mar 2010 21:09 On Mar 5, 6:34 pm, Gary Herron <gher...(a)islandtraining.com> wrote: > Mensanator wrote: > > On Mar 5, 3:42 pm, Gary Herron <gher...(a)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 -
From: Gary Herron on 5 Mar 2010 22:05 Mensanator wrote: > On Mar 5, 6:34 pm, Gary Herron <gher...(a)islandtraining.com> wrote: > >> Mensanator wrote: >> >>> On Mar 5, 3:42 pm, Gary Herron <gher...(a)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? > Not to my knowledge... I believe your statement is true in all cases which explicitly state the stop value. > >> 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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