def proper_divisors(n):
    """
    Return the sum of the proper divisors of positive integer n
    """
    return sum([x for x in range(1,n) if int(n/x) == n/x])

The list comprehension is this function is inefficient in that it computes
n/x twice. I'd like to do an  a = n/x and use a in
"if int(a) == a", but I don't know how.

Thanks,

Dick Moores
_______________________________________________
Tutor maillist  -  Tutor@python.org
To unsubscribe or change subscription options:
http://mail.python.org/mailman/listinfo/tutor

Reply via email to