For prime number generation I always use the Sieve of Eratosthenes:

def primes(n):
if n<=2:
return []
sieve=[True]*(n+1)
for x in range(3,int(n**0.5)+1,2):
for y in range(3,(n//x)+1,2):
sieve[(x*y)]=False
         

return [2]+[i for i in range(3,n,2) if sieve[i]]


In [42]: %timeit primes(10**5)
10 loops, best of 3: 60.4 ms per loop


In [43]: %timeit primes(10**6)
1 loops, best of 3: 1.01 s per loop

You can use Miller-Rabin primality test to check whether a number is prime or not. You can find its Python implementations here.

Always use timeit module to time your code, the 2nd one takes just 15us:

def func():
n = 600851475143
i = 2
while i * i < n:
while n % i == 0:
n = n / i
i = i + 1


In [19]: %timeit func()
1000 loops, best of 3: 1.35 ms per loop


def func():
i=1
while i<100:i+=1
....:


In [21]: %timeit func()
10000 loops, best of 3: 15.3 us per loop

Ok. So you said you understand the basics, but you're not sure EXACTLY how it works. First of all, this is a great answer to the Project Euler question it stems from. I've done a lot of research into this problem and this is by far the simplest response.

For the purpose of explanation, I'll let n = 20. To run the real Project Euler problem, let n = 600851475143.

n = 20
i = 2


while i * i < n:
while n%i == 0:
n = n / i
i = i + 1


print (n)

This explanation uses two while loops. The biggest thing to remember about while loops is that they run until they are no longer true.

The outer loop states that while i * i isn't greater than n (because the largest prime factor will never be larger than the square root of n), add 1 to i after the inner loop runs.

The inner loop states that while i divides evenly into n, replace n with n divided by i. This loop runs continuously until it is no longer true. For n=20 and i=2, n is replaced by 10, then again by 5. Because n0 doesn't evenly divide into 5, the loop stops with n2 and the outer loop finishes, producing n3.

Finally, because 3 squared is greater than 5, the outer loop is no longer true and prints the result of n.

Thanks for posting this. I looked at the code forever before realizing how exactly it worked. Hopefully, this is what you're looking for in a response. If not, let me know and I can explain further.

The code is wrong with 100. It should check case i * i = n:

I think it should be:

while i * i <= n:
if i * i = n:
n = i
break


while n%i == 0:
n = n / i
i = i + 1


print (n)

You shouldn't loop till the square root of the number! It may be right some times, but not always!

Largest prime factor of 10 is 5, which is bigger than the sqrt(10) (3.16, aprox).

Largest prime factor of 33 is 11, which is bigger than the sqrt(33) (5.5,74, aprox).

You're confusing this with the propriety which states that, if a number has a prime factor bigger than its sqrt, it has to have at least another one other prime factor smaller than its sqrt. So, with you want to test if a number is prime, you only need to test till its sqrt.

Isn't largest prime factor of 27 is 3 ?? The above code might be fastest,but it fails on 27 right ? 27 = 3*3*3 The above code returns 1 As far as I know.....1 is neither prime nor composite

I think, this is the better code

def prime_factors(n):
factors=[]
d=2
while(d*d<=n):
while(n>1):
while n%d==0:
factors.append(d)
n=n/d
d+=1
return factors[-1]

This question was the first link that popped up when I googled "python prime factorization". As pointed out by @quangpn88, this algorithm is wrong (!) for perfect squares such as n = 4, 9, 16, ... However, @quangpn88's fix does not work either, since it will yield incorrect results if the largest prime factor occurs 3 or more times, e.g., n = 2*2*2 = 8 or n = 2*3*3*3 = 54.

I believe a correct, brute-force algorithm in Python is:

def largest_prime_factor(n):
i = 2
while i * i <= n:
if n % i:
i += 1
else:
n //= i
return n

Don't use this in performance code, but it's OK for quick tests with moderately large numbers:

In [1]: %timeit largest_prime_factor(600851475143)
1000 loops, best of 3: 388 µs per loop

If the complete prime factorization is sought, this is the brute-force algorithm:

def prime_factors(n):
i = 2
factors = []
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return factors

Another way of doing this:

import sys
n = int(sys.argv[1])
result = []
for i in xrange(2,n):
while n % i == 0:
#print i,"|",n
n = n/i
result.append(i)


if n == 1:
break


if n > 1: result.append(n)
print result

sample output :
python test.py 68
[2, 2, 17]

Another way that skips even numbers after 2 is handled:

def prime_factors(n):
factors = []
d    = 2
step = 1
while d*d <= n:
while n>1:
while n%d == 0:
factors.append(d)
n = n/d
d += step
step = 2


return factors

It looks like people are doing the Project Euler thing where you code the solution yourself. For everyone else who wants to get work done, there's the primefac module which does very large numbers very quickly:

#!python


import primefac
import sys


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

"""
The prime factors of 13195 are 5, 7, 13 and 29.


What is the largest prime factor of the number 600851475143 ?


"""


from sympy import primefactors
print(primefactors(600851475143)[-1])

def find_prime_facs(n):
list_of_factors=[]
i=2
while n>1:
if n%i==0:
list_of_factors.append(i)
n=n/i
i=i-1
i+=1
return list_of_factors

My code:

# METHOD: PRIME FACTORS
def prime_factors(n):
'''PRIME FACTORS: generates a list of prime factors for the number given
RETURNS: number(being factored), list(prime factors), count(how many loops to find factors, for optimization)
'''
num = n                         #number at the end
count = 0                       #optimization (to count iterations)
index = 0                       #index (to test)
t = [2, 3, 5, 7]                #list (to test)
f = []                          #prime factors list
while t[index] ** 2 <= n:
count += 1                  #increment (how many loops to find factors)
if len(t) == (index + 1):
t.append(t[-2] + 6)     #extend test list (as much as needed) [2, 3, 5, 7, 11, 13...]
if n % t[index]:            #if 0 does else (otherwise increments, or try next t[index])
index += 1              #increment index
else:
n = n // t[index]       #drop max number we are testing... (this should drastically shorten the loops)
f.append(t[index])      #append factor to list
if n > 1:
f.append(n)                 #add last factor...
return num, f, f'count optimization: {count}'

Which I compared to the code with the most votes, which was very fast

    def prime_factors2(n):
i = 2
factors = []
count = 0                           #added to test optimization
while i * i <= n:
count += 1                      #added to test optimization
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return factors, f'count: {count}'   #print with (count added)

TESTING, (note, I added a COUNT in each loop to test the optimization)

# >>> prime_factors2(600851475143)
# ([71, 839, 1471, 6857], 'count: 1472')
# >>> prime_factors(600851475143)
# (600851475143, [71, 839, 1471, 6857], 'count optimization: 494')

I figure this code could be modified easily to get the (largest factor) or whatever else is needed. I'm open to any questions, my goal is to improve this much more as well for larger primes and factors.

In case you want to use numpy here's a way to create an array of all primes not greater than n:

[ i for i in np.arange(2,n+1) if 0 not in np.array([i] * (i-2) ) % np.arange(2,i)]

Below are two ways to generate prime factors of given number efficiently:

from math import sqrt




def prime_factors(num):
'''
This function collectes all prime factors of given number and prints them.
'''
prime_factors_list = []
while num % 2 == 0:
prime_factors_list.append(2)
num /= 2
for i in range(3, int(sqrt(num))+1, 2):
if num % i == 0:
prime_factors_list.append(i)
num /= i
if num > 2:
prime_factors_list.append(int(num))
print(sorted(prime_factors_list))




val = int(input('Enter number:'))
prime_factors(val)




def prime_factors_generator(num):
'''
This function creates a generator for prime factors of given number and generates the factors until user asks for them.
It handles StopIteration if generator exhausted.
'''
while num % 2 == 0:
yield 2
num /= 2
for i in range(3, int(sqrt(num))+1, 2):
if num % i == 0:
yield i
num /= i
if num > 2:
yield int(num)




val = int(input('Enter number:'))
prime_gen = prime_factors_generator(val)
while True:
try:
print(next(prime_gen))
except StopIteration:
print('Generator exhausted...')
break
else:
flag = input('Do you want next prime factor ? "y" or "n":')
if flag == 'y':
continue
elif flag == 'n':
break
else:
print('Please try again and enter a correct choice i.e. either y or n')

def prime(n):
for i in range(2,n):
if n%i==0:
return False
return True


def primefactors():
m=int(input('enter the number:'))
for i in range(2,m):
if (prime(i)):
if m%i==0:
print(i)
return print('end of it')


primefactors()

Check this out, it might help you a bit in your understanding.

#program to find the prime factors of a given number
import sympy as smp


try:
number = int(input('Enter a number : '))
except(ValueError) :
print('Please enter an integer !')
num = number
prime_factors = []
if smp.isprime(number) :
prime_factors.append(number)
else :
for i in range(2, int(number/2) + 1) :
"""while figuring out prime factors of a given number, n
keep in mind that a number can itself be prime or if not,
then all its prime factors will be less than or equal to its int(n/2 + 1)"""
if smp.isprime(i) and number % i == 0 :
while(number % i == 0) :
prime_factors.append(i)
number = number  / i
print('prime factors of ' + str(num) + ' - ')
for i in prime_factors :
print(i, end = ' ')

Since nobody has been trying to hack this with old nice reduce method, I'm going to take this occupation. This method isn't flexible for problems like this because it performs loop of repeated actions over array of arguments and there's no way how to interrupt this loop by default. The door open after we have implemented our own interupted reduce for interrupted loops like this:

from functools import reduce


def inner_func(func, cond, x, y):
res = func(x, y)
if not cond(res):
raise StopIteration(x, y)
return res


def ireducewhile(func, cond, iterable):
# generates intermediary results of args while reducing
iterable = iter(iterable)
x = next(iterable)
yield x
for y in iterable:
try:
x = inner_func(func, cond, x, y)
except StopIteration:
break
yield x

After that we are able to use some func that is the same as an input of standard Python reduce method. Let this func be defined in a following way:

def division(c):
num, start = c
for i in range(start, int(num**0.5)+1):
if num % i == 0:
return (num//i, i)
return None

Assuming we want to factor a number 600851475143, an expected output of this function after repeated use of this function should be this:

(600851475143, 2) -> (8462696833 -> 71), (10086647 -> 839), (6857, 1471) -> None

The first item of tuple is a number that division method takes and tries to divide by the smallest divisor starting from second item and finishing with square root of this number. If no divisor exists, None is returned. Now we need to start with iterator defined like this:

def gener(prime):
# returns and infinite generator (600851475143, 2), 0, 0, 0...
yield (prime, 2)
while True:
yield 0

Finally, the result of looping is:

result = list(ireducewhile(lambda x,y: div(x), lambda x: x is not None, iterable=gen(600851475143)))
#result: [(600851475143, 2), (8462696833, 71), (10086647, 839), (6857, 1471)]

And outputting prime divisors can be captured by:

if len(result) == 1: output = result[0][0]
else: output = list(map(lambda x: x[1], result[1:]))+[result[-1][0]]
#output: [2, 71, 839, 1471]

Note:

In order to make it more efficient, you might like to use pregenerated primes that lies in specific range instead of all the values of this range.

This is my python code: it has a fast check for primes and checks from highest to lowest the prime factors. You have to stop if no new numbers came out. (Any ideas on this?)

import math




def is_prime_v3(n):
""" Return 'true' if n is a prime number, 'False' otherwise """
if n == 1:
return False


if n > 2 and n % 2 == 0:
return False


max_divisor = math.floor(math.sqrt(n))
for d in range(3, 1 + max_divisor, 2):
if n % d == 0:
return False
return True




number = <Number>


for i in range(1,math.floor(number/2)):
if is_prime_v3(i):
if number % i == 0:
print("Found: {} with factor {}".format(number / i, i))

The answer for the initial question arrives in a fraction of a second.

If you are looking for pre-written code that is well maintained, use the function sympy.ntheory.primefactors from SymPy.

It returns a sorted list of prime factors of n.

>>> from sympy.ntheory import primefactors
>>> primefactors(6008)
[2, 751]

Pass the list to max() to get the biggest prime factor: max(primefactors(6008))

In case you want the prime factors of n and also the multiplicities of each of them, use sympy.ntheory.factorint.

Given a positive integer n, factorint(n) returns a dict containing the prime factors of n as keys and their respective multiplicities as values.

>>> from sympy.ntheory import factorint
>>> factorint(6008)   # 6008 = (2**3) * (751**1)
{2: 3, 751: 1}

The code is tested against Python 3.6.9 and SymPy 1.1.1.