On Mon, Nov 8, 2010 at 12:36 AM, Richard D. Moores <rdmoo...@gmail.com> wrote: > 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. >
You can't do that inside a list comprehension. Either get rid of the comprehension and do a regular loop, or get rid of the n/x expression. I'd suggest replacing the whole check with x % n == 0. n is a proper divisor of x if there is no remainder after division, after all. This also means you won't have to do a cast, which tend to be fairly expensive. On another note, getting rid of the list comprehension and using a generator expression will be even faster, since you won't have to build the list. Hugo _______________________________________________ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
