Python: Returning every element from a list in a normal-looking way for integer factorization function [duplicate]

Question

I'm a college student practicing some coding before school starts back up and I made a python function that is supposed to give the prime factorization of a number (excluding 1). It references a function that I made to check whether a number is prime, so I've included that function as well, although it shouldn't be important to the problem.

#Prime number checker

import math 
def is_prime(n):
    assert n >= 2
    sqrt_n = math.ceil(math.sqrt(n))
    last_digit = n % 10
    #First, incorporate special cases
    if last_digit == 0 or last_digit == 2 or last_digit == 4 or last_digit == 6 or last_digit == 8:
        return False
    #if the number is divisible by any of the prime numbers in this range, then it is not prime
    if n > 5 and last_digit == 5:
        return False
    appropriate_list = [i for i in range(2, sqrt_n + 1)]
    if any(n % i == 0 for i in appropriate_list):
        return False
    return True
    
#The integer factorization function
def integer_factorization(n):
    factors_list = []
    if is_prime(n) == True:
        output_list = []
        if factors_list: 
            output_list = [x for i in factors_list for x in i]
            return output_list, n
        else:
            return n
    else:
        divisors_list = [2, 3, 5, 7]
        for i in divisors_list:
            if n % i == 0:
                factors_list.append(i)
                return integer_factorization(n // i), factors_list

My problem: I want the output to look normal, e.g. for the output of integer_factorization(20) to just be 5, 2, 2 in whatever order. Instead, the output looks like a series of lists within lists. E.g. integer_factorization(20) results in this: ((5, [2]), [2])

I am running these programs on Jupyter notebook.

Can anyone assist with this?

because in some of your return statements, you return tuples: return output_list, n, and here: return integer_factorization(n // i), factors_list, if that isn't what you want, then don't return tuples — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Aug 10, 2023 at 21:53
I don't understand how this works at all. When the number is prime, factors_list is empty, so output_list will also be empty. The loop that appends to factors_list returns immediately after the first recursive call, so it stops at the first prime factor. — Barmar
– Barmar, Commented Aug 10, 2023 at 21:55

Kevin Eaverquepedo · Accepted Answer · 2023-08-10 21:56:38Z

Your issues:

return output_list, n this returns a tuple, a list and an integer, e.g. ([2,3], 2)

Kind of the same thing here return integer_factorization(n // i), factors_list this returns a tuple of two lists.

Try this:

#Prime number checker

import math
def is_prime(n):
    assert n >= 2
    sqrt_n = math.ceil(math.sqrt(n))
    last_digit = n % 10
    #First, incorporate special cases
    if last_digit == 0 or last_digit == 2 or last_digit == 4 or last_digit == 6 or last_digit == 8:
        return False
    #if the number is divisible by any of the prime numbers in this range, then it is not prime
    if n > 5 and last_digit == 5:
        return False
    appropriate_list = [i for i in range(2, sqrt_n + 1)]
    if any(n % i == 0 for i in appropriate_list):
        return False
    return True

#The integer factorization function
def integer_factorization(n):
    factors_list = []
    if is_prime(n) == True:
        output_list = []
        if factors_list:
            output_list = [x for i in factors_list for x in i]
            output_list.append(n)
            return output_list
        else:
            return [n]
    else:
        divisors_list = [2, 3, 5, 7]
        for i in divisors_list:
            if n % i == 0:
                factors_list.append(i)
                new_factors = integer_factorization(n // i)
                new_factors.extend(factors_list)
                return new_factors

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