a little bit of recursion

[Dominik Kaspar]

I'm trying to write a quite big recursive function, but the baseline
is the following:

def rek(N, L):
N = N + 1
if N >= 9: return L
else: return rek(N, L = L.append(N))

print rek(0, [])

Why isn't this working?
In my opinion it should return [1, 2, ..., 9]

Thanks for any answers.
Dominik

[rzed]

Re: a little bit of recursion

Dominik Kaspar wrote:[color=blue]
> I'm trying to write a quite big recursive function, but the baseline
> is the following:
>
> def rek(N, L):
> N = N + 1
> if N >= 9: return L
> else: return rek(N, L = L.append(N))
>
> print rek(0, [])
>
> Why isn't this working?
> In my opinion it should return [1, 2, ..., 9]
>
> Thanks for any answers.[/color]


One problem is that L = L.append(N) does not have the effect you want.
Try it on its own.
Another is that it won't return [1,2,...,9] even when that is
corrected.

[Skip Montanaro]

Re: a little bit of recursion


Dominik> def rek(N, L):
Dominik> N = N + 1
Dominik> if N >= 9: return L
Dominik> else: return rek(N, L = L.append(N))

In the recursive calls, L is None. Just pass L as the second arg:

def rek(N, L):
N = N + 1
if N >= 9:
return L
L.append(N)
return rek(N, L)

Skip

[Gregor Lingl]

Re: a little bit of recursion



rzed schrieb:[color=blue]
> Dominik Kaspar wrote:
>[color=green]
>>I'm trying to write a quite big recursive function, but the baseline
>>is the following:
>>
>>def rek(N, L):
>> N = N + 1
>> if N >= 9: return L
>> else: return rek(N, L = L.append(N))
>>
>>print rek(0, [])
>>
>>Why isn't this working?
>>In my opinion it should return [1, 2, ..., 9]
>>
>>Thanks for any answers.[/color]
>
>
>
> One problem is that L = L.append(N) does not have the effect you want.
> Try it on its own.
> Another is that it won't return [1,2,...,9] even when that is
> corrected.[/color]

That certainly depends on the way one corrects it.
However Dominik first has to notice, that append is a
method, which *changes* L but doesn't return L. Instead
it returns None. So it may be ok to call L.append(N),
but not on the right side of an assignment statement.

You can obsrve what I mean here:
[color=blue][color=green][color=darkred]
>>> L = [1,2,3]
>>> print L.append(4)[/color][/color][/color]
None[color=blue][color=green][color=darkred]
>>> L[/color][/color][/color]
[1, 2, 3, 4][color=blue][color=green][color=darkred]
>>>[/color][/color][/color]

HTH, Gregor[color=blue]
>
>
>
>[/color]

[Gregor Lingl]

Re: a little bit of recursion



Skip Montanaro schrieb:[color=blue]
> Dominik> def rek(N, L):
> Dominik> N = N + 1
> Dominik> if N >= 9: return L
> Dominik> else: return rek(N, L = L.append(N))
>
> In the recursive calls, L is None. Just pass L as the second arg:
>
> def rek(N, L):
> N = N + 1
> if N >= 9:
> return L
> L.append(N)
> return rek(N, L)
>
> Skip
>[/color]
Nice, but will not append 9 as desired
Gregor[color=blue]
>[/color]

[Gregor Lingl]

Re: a little bit of recursion



Skip Montanaro schrieb:[color=blue]
> Dominik> def rek(N, L):
> Dominik> N = N + 1
> Dominik> if N >= 9: return L
> Dominik> else: return rek(N, L = L.append(N))
>
> In the recursive calls, L is None. Just pass L as the second arg:
>
> def rek(N, L):
> N = N + 1
> if N >= 9:
> return L
> L.append(N)
> return rek(N, L)
>
> Skip
>[/color]
Nice, but will not append 9 as desired
Gregor[color=blue]
>[/color]

[Skip Montanaro]

Re: a little bit of recursion


Gregor> Skip Montanaro schrieb:
Dominik> def rek(N, L):
Dominik> N = N + 1
Dominik> if N >= 9: return L
Dominik> else: return rek(N, L = L.append(N))
[color=blue][color=green]
>> In the recursive calls, L is None. Just pass L as the second arg:
>>
>> def rek(N, L):
>> N = N + 1
>> if N >= 9:
>> return L
>> L.append(N)
>> return rek(N, L)[/color][/color]

Gregor> Nice, but will not append 9 as desired

Sure, but that's a logic error, not a misunderstandin g about what L.append()
returns. I suspect the latter was what confused Dominik. Once he runs
something that doesn't raise an AttributeError, he'll quickly discover that
his test should be "if N > 9". He couldn't even get to that point with his
original code.

Skip

[Georgy Pruss]

Re: a little bit of recursion


"Gregor Lingl" <glingl@aon.a t> wrote in message news:3F95A439.4 060102@aon.at.. .[color=blue]
>[color=green]
> > <...>
> >
> > def rek(N, L):
> > N = N + 1
> > if N >= 9:
> > return L
> > L.append(N)
> > return rek(N, L)
> >
> > Skip
> >[/color]
> Nice, but will not append 9 as desired
> Gregor[color=green]
> >[/color]
>[/color]

def rek(N, L):
if N >= 9: return L
N += 1
L.append(N)
return rek(N,L)

G-:

--
Georgy Pruss
E^mail: 'ZDAwMTEyMHQwMz MwQGhvdG1haWwuY 29t\n'.decode(' base64')