# [Tutor] List comprehension question

Stefan Behnel stefan_ml at behnel.de
Tue Nov 9 12:35:26 CET 2010

Richard D. Moores, 09.11.2010 12:07:
>>>> That sqrt(n) works for speeding up the finding of primes, but here we
>>>> want to use int(n/2) (and why didn't I think of that?), which results
>>>> in about a 2x speedup. See<http://tutoree7.pastebin.com/dyRC8vuX>.
>>>
>>> NO! Use  int(n/2)+1 .  I'll correct that in
>>> <http://tutoree7.pastebin.com/dyRC8vuX>  and report back.
>>
>> See<http://tutoree7.pastebin.com/eZPaVpwG>
>>
>> But now I'll have to redo again using Stefan's ingenious suggestion.
>
> Here's what I wrote using Stefan's suggestion:
>
> def proper_divisors_sum(n):
>      pd_list = []
>      for x in range(1, int(n**.5)+1):
>          if n % x == 0:
>              factor = int(x)
>              factor2 = int(n/factor)
>              pd_list.extend([factor, factor2])
>      pd_list = list(set(pd_list))
>      pd_list.sort()
>      pd_list = pd_list[:-1]
>      return sum(pd_list)

Have a look at divmod():

http://docs.python.org/library/functions.html#divmod

Note that you don't need to convert 'x' to an int. It already has that type
(see the docs on range()). Also, int(n/factor) is the same as n//factor here.

Stefan

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