Python recursive program for simple factorization of a number

Question

I wrote the following program for simple factorization of a number:

import math
def prime_factorize(x,li=[]):
    until = int(math.sqrt(x))+1
    for i in xrange(2,until):
        if not x%i:
            li.append(i)
            break
    else:                      #This else belongs to for
        li.append(x)
        print li               #First print statement; This is what is returned
        return li
    prime_factorize(x/i,li)

if __name__=='__main__':
    print prime_factorize(300)   #Second print statement, WTF. why is this None

The following is the conclusion:

 [2, 2, 3, 5, 5]
 None

Altho ', the return value is printed correctly, after the return value is printed all the time. What am I missing?

Also, how can I improve the program (while continuing to use recursion)

Answer 1

prime_factorize return - "return prime_factorize (x/i, li)" . ( ), , .

, , - :

def prime_factorize(x,li=None):
    if li is None: li = []

:

>>> prime_factorize(10)
[2, 5]
>>> prime_factorize(4)
[2, 5, 2, 2]
>>> prime_factorize(19)
[2, 5, 2, 2, 19]

Answer 2

, , . , .

def primeFact (i, f):
    if i < f:
        return []
    if i % f == 0:
        return [f] + primeFact (i / f, 2)
    return primeFact (i, f + 1)

.

>>> primeFact (300, 2)
[2, 2, 3, 5, 5]
>>> primeFact (17, 2)
[17]
>>> primeFact (2310, 2)
[2, 3, 5, 7, 11]

Answer 3

@ print. , , , , :

def prime_factorize(x):
  li = []
  while x >= 2:
    until = int(math.sqrt(x))+1
    for i in xrange(2,until):
      if not x%i:
        li.append(i)
        break
    else:
      li.append(x)
      return li
    x //= i

( - , ), Python , .

" [non-base-case] 'return thisfun(newargs)' args=newargs; continue while True:" - . li non-arg ( arg), while continue, .

(sqrt avoidance, memoization,...) .

Answer 4

More functional version.

def prime_factorize( number ):
    def recurse( factors, x, n ):
        if x<2: return factors # 0,1 dont have prime factors
        if n > 1+x**0.5: # reached the upper limit
            factors.append( x ) # the only prime left is x itself
            return factors
        if x%n==0: # x is a factor
            factors.append( n )
            return recurse( factors, x/n, n )
        else:
            return recurse( factors, x, n+1 )
    return recurse( [], number, 2)

for num, factors in ((n, prime_factorize( n )) for n in range(1,50000)):
    assert (num==reduce(lambda x,y:x*y, factors, 1)), (num, factors)
    #print num, ":", factors

Answer 5

def primeFactorization(n):
    """ Return the prime factors of the given number. """
    factors = []
    lastresult = n
    while 1:
        if lastresult == 1:
            break

        c = 2

        while 1:
            if lastresult % c == 0:
                break

            c += 1

        factors.append(c)
        lastresult /= c

    return factors

it's good.