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]

Answer 2

def factorize(n):
  for f in range(2,n//2+1):
    while n%f == 0:
      n //= f
      yield f

It's slow but dead simple. If you want to create a command-line utility, you could do:

import sys
[print(i) for i in factorize(int(sys.argv[1]))]

Answer 3

Simple way to get the desired solution

def Factor(n):
    d = 2
    factors = []
    while n >= d*d:
        if n % d == 0:
            n//=d
            # print(d,end = " ")
            factors.append(d)
        else:
            d = d+1
    if n>1:
        # print(int(n))
        factors.append(n)
    return factors

Answer 4

def get_prime_factors(number):
    """
    Return prime factor list for a given number
        number - an integer number
        Example: get_prime_factors(8) --> [2, 2, 2].
    """
    if number == 1:
        return []

    # We have to begin with 2 instead of 1 or 0
    # to avoid the calls infinite or the division by 0
    for i in xrange(2, number):
        # Get remainder and quotient
        rd, qt = divmod(number, i)
        if not qt: # if equal to zero
            return [i] + get_prime_factors(rd)

    return [number]

Answer 5

def prime_factors(num, dd=2):
    while dd <= num and num>1:
        if num % dd == 0:
            num //= dd
            yield dd
        dd +=1

Lot of answers above fail on small primes, e.g. 3, 5 and 7. The above is succinct and fast enough for ordinary use.

print list(prime_factors(3))

[3]