I have written a code to find out the LCM (Lowest Common Multiple) of a list of numbers but there appears to be an error in my code. The code is given below:

def final_lcm(thelist):
   previous_thelist = thelist
   prime_thelist = list(set(thelist) - set(returns_new_thelist(previous_thelist))
   factors = 1
   for i in prime_thelist:
       factors = factors*i
   new_thelist = returns_new_thelist(previous_thelist)
   for i in range(1, 10000000000):
       s_empty = []
       for j in new_thelist:
           if i % j  == 0:
               s_empty.append(True)
       if len(new_thelist) == len(s_empty):
           initial_lcm = i
           break
   final_lcm = factor*initial_lcm
   return final_lcm



def returns_new_thelist(ll):
    if 3 in ll:
        ll.remove(3)
    for i in ll:
        if checks_if_prime(i) == True:
            ll.remove(i)
    return ll    

def checks_if_prime(n):
    if n == 2:
    return True
    import math
    for i in range(math.ceil(0.5*n), 1, -1):
        if n % i == 0:
            return False
        elif i == 2:
            return True

print(final_lcm([1,2,3,4,5,6,7,8,9]))

Kindly pardon my poor choice of variables, I request you to see if the logic is correct and that the code is functional.

The syntax error which I am getting is that "factors" is invalid syntax though I don't agree with this. Please tell me where my code is wrong.

As of Python 3.9 lcm() function has been added in the math library. It can be called with the following signature:

math.lcm(*integers)

Return the least common multiple of the specified integer arguments. If all arguments are nonzero, then the returned value is the smallest positive integer that is a multiple of all arguments. If any of the arguments is zero, then the returned value is 0. lcm() without arguments returns 1.

Advantages:

1. Besides being native,
2. Its a one-liner,
3. Its fastest,
4. Can deal with arbitrarily long list of integers
5. And can deal with nearly any kind of exceptions (e.g. lcm(0,0)) overlooked by custom-built solutions.