Need to make a list by splitting the given string in python

Question

Input string has 'n' times of X. Like this "XXXXX" Need to split the string and make out a list in which each character of the string(X) is each element of list.

Constraints:
0 <= X <= any number
'n' is unknown

Example 1: X is 12,

Input string : "121212121212121212"
Output list : [12,12,12,12,12,12,12,12,12]

Example 2: X is 2,

Input string : "2222"
Output list : [2,2,2,2]

Example 3: X is 489,

Input string : "489489489"
Output list : [489,489,489]

I tried this approach my_list = list(input_string)

I am getting output like, [1,2,1,2,1,2,1,2,1,2,1,2,1,2,1,2,1,2]

Can anyone help me to resolve this?

Edit 1: I am new to stackoverflow. Please help me to edit this question properly.

Edit 2: As this question is clear and understandable, please consider upvoting.

Answer 1

To decode this string you need 'n' and X anyway. Why not build it directly with:

li = [X] * n

Edit: If you don't have 'n', you could get it easily from the given string. (Under the condition that givenString only consist of X)

n = len(givenString) / len(str(X))
li = [X] * n

Answer 2

You can use regular expressions:

import re 

x = input('Enter Your X : ')
mystr = input('Enter Your Input String : ')

result = re.findall(x, mystr) 
print(result)

This outputs:

Enter Your X : 12
Enter Your Input String : 121212121212
['12', '12', '12', '12', '12', '12']

Answer 3

Here's a variant that makes sure the substrings all match the given pattern:

input_ = "121212121212121212"
x = "12"
res = []
while input_.startswith(x):
    res.append(x)
    input_ = input_[len(x):]

Answer 4

If X is also unknown, you can use this:

import textwrap

def find_shortest_pattern(input_):
     for len_ in range(1, int(len(input_) / 2) + 1):
          patterns = textwrap.wrap(input_, len_)
          if all(pattern == patterns[0] for pattern in patterns):
               return patterns[0]

find_shortest_pattern("123123123")