how to find all the prime factors of a number in python code example

Example 1: prime factorization python

import math


def primeFactors(n):
    # no of even divisibility
    while n % 2 == 0:
        print(2)
        n = n / 2
    # n reduces to become odd
    for i in range(3, int(math.sqrt(n)) + 1, 2):
        # while i divides n
        while n % i == 0:
            print(i)
            n = n / i
    # if n is a prime
    if n > 2:
        print(n)


primeFactors(256)

Example 2: to find factors of a number in python

# Python Program to find the factors of a number

# This function computes the factor of the argument passed
def print_factors(x):
   print("The factors of",x,"are:")
   for i in range(1, x + 1):
       if x % i == 0:
           print(i)

num = 6

print_factors(num)

Example 3: python find factors of a number

def factors(n):
    return set(reduce(list.__add__, \
        ([i, n//i] for i in range(1, int(n**0.5) + 1) if not n % i )))

Example 4: python prime factors

# There is no quick way to calculate the prime factors of a number.
# In fact, prime factorization is so famously hard that it's what puts the "asymmetric" in asymmetric RSA encryption.
# That being said, it can be sped up a little bit by using divisibility rules, like checking if the sum of the digits is divisible by 3.

def factors(num):
        ps = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149] # Primes from https://primes.utm.edu/lists/small/10000.txt. Primes can also be generated by iterating through numbers and checking for factors, or by using a probabilistic test like Rabin-Miller.
        pdict = {}
        for p in ps:
                if p <= num:
                        while (num / p).is_integer():
                                if str(p) in pdict:
                                        pdict[str(p)] += 1
                                else:
                                        pdict[str(p)] = 1
                                num /= p
                if num == 1: break
        return pdict

# Returns a dictionary in the form {"base": "exponent"}