Question about generators

[Piet van Oostrum]

>>>>> Cameron Pulsford <cameron.pulsford at gmail.com> (CP) wrote:

>CP> Hey everyone, I have this small piece of code that simply finds the
>CP> factors of a number.

Others have already given you advice to add the [2, 3, 5] to the
iterator (of which the primes.extend([2,3,5]) will not work). Please
allow me to make some other remarks.

>CP> import sys

>CP> def factor(n):
>CP>     primes = (6*i+j for i in xrange(1, n) for j in [1, 5] if (i+j)%5 ! = 0)

It is a bit misleading to call this `primes' as it will also contain
non-primes. Of course they don't harm as they will never be considered a
factor because all their prime factors will have been successfully
removed before this non prime will be considered.

>CP>     factors = []

>CP>     for i in [2, 3, 5]:
>CP>         while n % i == 0:
>CP>             n /= i
>CP>             factors.append(i)
>CP>     for i in primes:
>CP>         while n % i == 0:
>CP>             n /= i
>CP>             factors.append(i)

         You can stop the loop when n == 1. Like:
                 if n == 1: break

>CP>     print factors

>CP> factor(int(sys.argv[1]))

>CP> My question is, is it possible to combine those two loops? The primes
>CP> generator I wrote finds all primes up to n, except for 2, 3 and 5, so  I
>CP> must check those explicitly. Is there anyway to concatenate the hard  coded
>CP> list of [2,3,5] and the generator I wrote so that I don't need  two for
>CP> loops that do the same thing?

>CP> I tried writing a primes function using yield statements, but it  didn't
>CP> work like I thought it would.

See the recent thread `Why is my code faster with append() in a loop than with a large list?'
http://mail.python.org/pipermail/python-list/2009-July/718931.html
-- 
Piet van Oostrum <piet at cs.uu.nl>
URL: http://pietvanoostrum.com [PGP 8DAE142BE17999C4]
Private email: piet at vanoostrum.org