> On Jul 30, 5:48 pm, beginner <zyzhu2... at gmail.com> wrote:
>
>> def f(n):
>> l=[]
>> while n>0:
>> l.append(n%26)
>> n /=26
>> return l
>>
>> I am wondering what is the 'functional' way to do the same.
>>
> I see. It is interesting (and not surprisingly) that recursion or
> yield are required. Thanks for everyone's help.
Why not use both recursion and generators?
def f2(n):
if n > 0:
n2, x = divmod(n, 26)
yield x
for next in f2(n2):
yield next
Could this be considered functional?
Anyway, I wonder if there is a more elegant way to do the "yield x... yield
next" part. Any ideas?
arnau
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