On Wednesday 02 March 2005 06:03, actuary77 wrote:
> It now makes sense if I write it, (simple):
>
> def rec2(n):
>      if n == 0:
>          return []
>      else:
>          return [n] + rec2(n-1)

Or, if you're not interested in a recursive function to do this job (which 
should be way faster...):

>>> def iter1(n):
...     nl = range(1,n+1)
...     nl.reverse()
...     return nl
...
>>> print iter1(4)
[4, 3, 2, 1]

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