[Tutor] List comprehension question

[Martin A. Brown]

Hello,

 : def proper_divisors_sum(n):
 :     pd_list = []
 :     for x in range(1, int(n**.5)+1):
 :         if n % x == 0:
 :             factor = int(x)
 :             pd_list.append(factor)
 :             factor2 = int(n/factor)
 :             pd_list.append(factor2)
 :     pd_list = list(set(pd_list))
 :     pd_list.sort()
 :     pd_list = pd_list[:-1]
 :     return sum(pd_list)

A few questions--noting, of course, that I'm not reading this with 
an eye toward performance, which it seems you are, but these occur 
to me:

  * Why use a list (pd_list = []), when you want unique divisors?
    Why not, pd = set() ?  Then, instead of pd_list.append(factor), 
    pd.add(factor) or something like pd.update((factor,factor2))?
    (See also my suggestion below.)

  * Also, since you are throwing away the divisor n, anyway, 
    why not skip it from the beginning?  Like this:

      pd_list = [1,]     
      for x in range(2, int(n**.5)+1):

  * So, I'm not terribly math aware, but I don't think I have 
    substantially changed the meaning of your program.  I find 
    breaking out the functions makes things a bit clearer to 
    somebody who might be reading my code later.

Combining these suggestions and splitting into separate functions 
(you don't really need to sort before summing), I end up with the 
below.  Note, that I stuff the proper divisor 1 in the set at 
creation time.  This is probably not worth it from an optimization 
perspective, but as I have learned, the only way to know is to time 
it (and I didn't time it).

      def proper_divisors_sum(n):
          return sum(proper_divisors(n))
      
      def proper_divisors_sorted(n):
          return sorted(proper_divisors(n))
      
      def proper_divisors(n):
          pd = set((1,))
          for x in range(2, int(n**.5)+1):
              factor, mod = divmod(n,x)
              if mod == 0:
                  pd.update((x,factor))
          return list(pd)

Have a great day,

-Martin

-- 
Martin A. Brown
http://linux-ip.net/