Prime factorization - list

Question

I am trying to implement a function primeFac() that takes as input a positive integer n and returns a list containing all the numbers in the prime

Answer 1

You can use sieve Of Eratosthenes to generate all the primes up to (n/2) + 1 and then use a list comprehension to get all the prime factors:

def rwh_primes2(n):
    # http://stackoverflow.com/questions/2068372/fastest-way-to-list-all-primes-below-n-in-python/3035188#3035188
    """ Input n>=6, Returns a list of primes, 2 <= p < n """
    correction = (n%6>1)
    n = {0:n,1:n-1,2:n+4,3:n+3,4:n+2,5:n+1}[n%6]
    sieve = [True] * (n/3)
    sieve[0] = False
    for i in xrange(int(n**0.5)/3+1):
      if sieve[i]:
        k=3*i+1|1
        sieve[      ((k*k)/3)      ::2*k]=[False]*((n/6-(k*k)/6-1)/k+1)
        sieve[(k*k+4*k-2*k*(i&1))/3::2*k]=[False]*((n/6-(k*k+4*k-2*k*(i&1))/6-1)/k+1)
    return [2,3] + [3*i+1|1 for i in xrange(1,n/3-correction) if sieve[i]]

def primeFacs(n):
    primes = rwh_primes2((n/2)+1)
    return [x for x in primes if n%x == 0]

print primeFacs(99999)
#[3, 41, 271]