You can use itertools:

from itertools import chain

a = 1
b = [2,3]
c = [4,5,6]
final_list = list(chain.from_iterable([[a], b, c]))

Output:

[1, 2, 3, 4, 5, 6]

However, if you do not know the contents of a, b, and c ahead of time, you can try this:

new_list = [[i] if not isinstance(i, list) else i for i in [a, b, c]]
final_list = list(chain.from_iterable(new_list))

There are already great methods already mentioned on this page, however it does seem a little obscure as to which to use. So I have added some mesurements so you can more easily decide for yourself:

A large number has been used (for overhead) 1111111111111122222222222222222333333333333333333333

Using map(int, str(num)):

import timeit

def method():
    num = 1111111111111122222222222222222333333333333333333333
    return map(int, str(num))

print(timeit.timeit("method()", setup="from __main__ import method", number=10000)

Output: 0.018631496999999997

Using list comprehension:

import timeit

def method():
    num = 1111111111111122222222222222222333333333333333333333
    return [int(x) for x in str(num)]

print(timeit.timeit("method()", setup="from __main__ import method", number=10000))

Output: 0.28403817900000006

Code taken from this answer

The results show that the first method involving inbuilt methods is much faster than list comprehension.

The "mathematical way":

import timeit

def method():
    q = 1111111111111122222222222222222333333333333333333333
    ret = []
    while q != 0:
        q, r = divmod(q, 10) # Divide by 10, see the remainder
        ret.insert(0, r) # The remainder is the first to the right digit
    return ret

print(timeit.timeit("method()", setup="from __main__ import method", number=10000))

Output: 0.38133582499999996

Code taken from this answer

The list(str(123)) method (does not provide the right output):

import timeit

def method():
    return list(str(1111111111111122222222222222222333333333333333333333))
    
print(timeit.timeit("method()", setup="from __main__ import method", number=10000))

Output: 0.028560138000000013

Code taken from this answer

The answer by Duberly González Molinari:

import timeit

def method():
    n = 1111111111111122222222222222222333333333333333333333
    l = []
    while n != 0:
        l = [n % 10] + l
        n = n // 10
    return l

print(timeit.timeit("method()", setup="from __main__ import method", number=10000))

Output: 0.37039988200000007

Code taken from this answer

Remarks:

In all cases the map(int, str(num)) is the fastest method (and is therefore probably the best method to use). List comprehension is the second fastest (but the method using map(int, str(num)) is probably the most desirable of the two.

Those that reinvent the wheel are interesting but are probably not so desirable in real use.

You should use raw_input and convert to int with a list comprehension:

user_input = raw_input('Enter the numbers: ')
my_list = [int(i) for i in user_input.split(',')]

From the offical documentation: raw_input reads a line from input, converts it to a string (stripping a trailing newline), and returns that.

Here:

a = 1234
lst = []
lst.append(a)

print(lst) #[1234]

mylist = [i] 

This will create a list called mylist with exactly one element i. This can be extended to create a list with as many values as you want: For example:

mylist = [i1,i2,i3,i4]

This creates a list with four element i1,i2,i3,i4 in that order. This is more efficient that appending each one of the elements to the list.