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 factorization of n.

I have gotten this far but I think it would be better to use recursion here, not sure how to create a recursive code here, what would be the base case? to start with.

My code:

def primes(n):
    primfac = []
    d = 2
    while (n > 1):
         if n%d==0:
             primfac.append(d)
    # how do I continue from here... ?

Answer 1

A simple trial division:

def primes(n):
    primfac = []
    d = 2
    while d*d <= n:
        while (n % d) == 0:
            primfac.append(d)  # supposing you want multiple factors repeated
            n //= d
        d += 1
    if n > 1:
       primfac.append(n)
    return primfac

with O(sqrt(n)) complexity (worst case). You can easily improve it by special-casing 2 and looping only over odd d (or special-casing more small primes and looping over fewer possible divisors).

Answer 2

The primefac module does factorizations with all the fancy techniques mathematicians have developed over the centuries:

#!python

import primefac
import sys

n = int( sys.argv[1] )
factors = list( primefac.primefac(n) )
print '\n'.join(map(str, factors))

Answer 3

This is a comprehension based solution, it might be the closest you can get to a recursive solution in Python while being possible to use for large numbers.

You can get proper divisors with one line:

divisors = [ d for d in xrange(2,int(math.sqrt(n))) if n % d == 0 ]

then we can test for a number in divisors to be prime:

def isprime(d): return all( d % od != 0 for od in divisors if od != d )

which tests that no other divisors divides d.

Then we can filter prime divisors:

prime_divisors = [ d for d in divisors if isprime(d) ]

Of course, it can be combined in a single function:

def primes(n):
    divisors = [ d for d in range(2,n//2+1) if n % d == 0 ]
    return [ d for d in divisors if \
             all( d % od != 0 for od in divisors if od != d ) ]

Here, the \ is there to break the line without messing with Python indentation.

Answer 4

I've tweaked @user448810's answer to use iterators from itertools (and python3.4, but it should be back-portable). The solution is about 15% faster.

import itertools

def factors(n):
    f = 2
    increments = itertools.chain([1,2,2], itertools.cycle([4,2,4,2,4,6,2,6]))
    for incr in increments:
        if f*f > n:
            break
        while n % f == 0:
            yield f
            n //= f
        f += incr
    if n > 1:
        yield n

Note that this returns an iterable, not a list. Wrap it in list() if that's what you want.

Answer 5

Here is my version of factorization by trial division, which incorporates the optimization of dividing only by two and the odd integers proposed by Daniel Fischer:

def factors(n):
    f, fs = 3, []
    while n % 2 == 0:
        fs.append(2)
        n /= 2
    while f * f <= n:
        while n % f == 0:
            fs.append(f)
            n /= f
        f += 2
    if n > 1: fs.append(n)
    return fs

An improvement on trial division by two and the odd numbers is wheel factorization, which uses a cyclic set of gaps between potential primes to greatly reduce the number of trial divisions. Here we use a 2,3,5-wheel:

def factors(n):
    gaps = [1,2,2,4,2,4,2,4,6,2,6]
    length, cycle = 11, 3
    f, fs, nxt = 2, [], 0
    while f * f <= n:
        while n % f == 0:
            fs.append(f)
            n /= f
        f += gaps[nxt]
        nxt += 1
        if nxt == length:
            nxt = cycle
    if n > 1: fs.append(n)
    return fs

Thus, print factors(13290059) will output [3119, 4261]. Factoring wheels have the same O(sqrt(n)) time complexity as normal trial division, but will be two or three times faster in practice.

I've done a lot of work with prime numbers at my blog. Please feel free to visit and study.

Answer 6

Most of the above solutions appear somewhat incomplete. A prime factorization would repeat each prime factor of the number (e.g. 9 = [3 3]).

Also, the above solutions could be written as lazy functions for implementation convenience.

The use sieve Of Eratosthenes to find primes to test is optimal, but; the above implementation used more memory than necessary.

I'm not certain if/how "wheel factorization" would be superior to applying only prime factors, for division tests of n.

While these solution are indeed helpful, I'd suggest the following two functions -

Function-1 :

def primes(n):
    if n < 2: return
    yield 2
    plist = [2]
    for i in range(3,n):
        test = True
        for j in plist:
            if j>n**0.5:
                break
            if i%j==0:
                test = False
                break
        if test:
            plist.append(i)
            yield i

Function-2 :

def pfactors(n):
    for p in primes(n):
        while n%p==0:
            yield p
            n=n//p
            if n==1: return

list(pfactors(99999))
[3, 3, 41, 271]

3*3*41*271
99999

list(pfactors(13290059))
[3119, 4261]

3119*4261
13290059

Answer 7

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 8

Most of the answer are making things too complex. We can do this

def prime_factors(n):
    num = []

    #add 2 to list or prime factors and remove all even numbers(like sieve of ertosthenes)
    while(n%2 == 0):
        num.append(2)
        n /= 2

    #divide by odd numbers and remove all of their multiples increment by 2 if no perfectlly devides add it
    for i in xrange(3, int(sqrt(n))+1, 2):
        while (n%i == 0):
            num.append(i)
            n /= i

    #if no is > 2 i.e no is a prime number that is only divisible by itself add it
    if n>2:
        num.append(n)

    print (num)

Algorithm from GeeksforGeeks

Answer 9

prime factors of a number:

def primefactors(x):
    factorlist=[]
    loop=2
    while loop<=x:
        if x%loop==0:
            x//=loop
            factorlist.append(loop)
        else:
            loop+=1
    return factorlist

x = int(input())
alist=primefactors(x)
print(alist)

You'll get the list. If you want to get the pairs of prime factors of a number try this: http://pythonplanet.blogspot.in/2015/09/list-of-all-unique-pairs-of-prime.html

Answer 10

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 11

Here is an efficient way to accomplish what you need:

def prime_factors(n): 
  l = []
  if n < 2: return l
  if n&1==0:
    l.append(2)
    while n&1==0: n>>=1
  i = 3
  m = int(math.sqrt(n))+1
  while i < m:
    if n%i==0:
      l.append(i)
      while n%i==0: n//=i
    i+= 2
    m = int(math.sqrt(n))+1
  if n>2: l.append(n)
  return l

prime_factors(198765430488765430290) = [2, 3, 5, 7, 11, 13, 19, 23, 3607, 3803, 52579]

Answer 12

    from sets import Set
    # this function generates all the possible factors of a required number x
    def factors_mult(X):
        L = []
        [L.append(i) for i in range(2,X) if X % i == 0]
        return L

    # this function generates list containing prime numbers upto the required number x 
    def prime_range(X):
        l = [2]
        for i in range(3,X+1):
            for j in range(2,i):
               if i % j == 0:
               break
            else:    
               l.append(i)
        return l

    # This function computes the intersection of the two lists by invoking Set from the sets module
    def prime_factors(X):
        y = Set(prime_range(X))
        z = Set(factors_mult(X))
        k = list(y & z)
        k = sorted(k)

        print "The prime factors of " + str(X) + " is ", k

    # for eg
    prime_factors(356)

Answer 13

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 14

This is the code I made. It works fine for numbers with small primes, but it takes a while for numbers with primes in the millions.

def pfactor(num):
div = 2
pflist = []
while div <= num:
    if num % div == 0:
        pflist.append(div)
        num /= div
    else:
        div += 1
# The stuff afterwards is just to convert the list of primes into an expression
pfex = ''
for item in list(set(pflist)):
    pfex += str(item) + '^' + str(pflist.count(item)) + ' * '
pfex = pfex[0:-3]
return pfex

Answer 15

I would like to share my code for finding the prime factors of number given input by the user:

a = int(input("Enter a number: "))

def prime(a):
    b = list()
    i = 1
    while i<=a:
        if a%i ==0 and i!=1 and i!=a:
            b.append(i)
        i+=1
    return b

c = list()
for x in prime(a):
    if len(prime(x)) == 0:
        c.append(x)

print(c)

Answer 16

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]

Answer 17

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]