How would I be able to take a string like 'aaaaaaaaaaaaaaaaaaaaaaa'
and split it into 4 length tuples like (aaaa,aaaa,aaaa)

Use textwrap.wrap:

>>> import textwrap
>>> s="aaaaaaaaaaaaaaaaaaaaaaa"
>>> textwrap.wrap(s, 4)
['aaaa', 'aaaa', 'aaaa', 'aaaa', 'aaaa', 'aaa']

Using list comprehension, generator expression:

>>> s="aaaaaaaaaaaaaaaaaaaaaaa"
>>> [s[i:i+4] for i in range(0, len(s), 4)]
['aaaa', 'aaaa', 'aaaa', 'aaaa', 'aaaa', 'aaa']

>>> tuple(s[i:i+4] for i in range(0, len(s), 4))
('aaaa', 'aaaa', 'aaaa', 'aaaa', 'aaaa', 'aaa')

>>> s="a bcdefghi j"
>>> tuple(s[i:i+4] for i in range(0, len(s), 4))
('a bc', 'defg', 'hi j')

Another solution using regex:

>>> s="aaaaaaaaaaaaaaaaaaaaaaa"
>>> import re
>>> re.findall('[a-z]{4}', s)
['aaaa', 'aaaa', 'aaaa', 'aaaa', 'aaaa']
>>>

You could use the grouper recipe, zip(*[iter(s)]*4):

In [113]: s="aaaaaaaaaaaaaaaaaaaaaaa"

In [114]: [''.join(item) for item in zip(*[iter(s)]*4)]
Out[114]: ['aaaa', 'aaaa', 'aaaa', 'aaaa', 'aaaa']

Note that textwrap.wrap may not split s into strings of length 4 if the string contains spaces:

In [43]: textwrap.wrap('I am a hat', 4)
Out[43]: ['I am', 'a', 'hat']

The grouper recipe is faster than using textwrap:

In [115]: import textwrap

In [116]: %timeit [''.join(item) for item in zip(*[iter(s)]*4)]
100000 loops, best of 3: 2.41 µs per loop

In [117]: %timeit textwrap.wrap(s, 4)
10000 loops, best of 3: 32.5 µs per loop

And the grouper recipe can work with any iterator, while textwrap only works with strings.

s="abcdef"

We need to split in parts of 2

[s[pos:pos+2] for pos,i in enumerate(list(s)) if pos%2 == 0]

Answer:

['ab', 'cd', 'ef']

s="abcdefghi"

k – no of parts of string

k = 3

parts – list to store parts of string

parts = [s[i:i+k] for i in range(0, len(s), k)]

parts –> [‘abc’, ‘def’, ‘ghi’]

I think this method is simpler. But the message length must be split with split_size. Or letters must be added to the message. Example: message = “lorem ipsum_” then the added letter can be deleted.

message = "lorem ipsum"

array = []

temp = ""

split_size = 3

for i in range(1, len(message) + 1):
    temp += message[i - 1]

    if i % split_size == 0:
        array.append(temp)
        temp = ""

print(array)

Output:
[‘lor’, ’em ‘, ‘ips’]

Here’s another possible solution to the given problem:

def split_by_length(text, width):
    width = max(1, width)
    chunk = ""
    for v in text:
        chunk += v
        if len(chunk) == width:
            yield chunk
            chunk = ""

    if chunk:
        yield chunk

if __name__ == '__main__':
    x = "123456789"
    for i in range(20):
        print(i, list(split_by_length(x, i)))

Output:

0 ['1', '2', '3', '4', '5', '6', '7', '8', '9']
1 ['1', '2', '3', '4', '5', '6', '7', '8', '9']
2 ['12', '34', '56', '78', '9']
3 ['123', '456', '789']
4 ['1234', '5678', '9']
5 ['12345', '6789']
6 ['123456', '789']
7 ['1234567', '89']
8 ['12345678', '9']
9 ['123456789']
10 ['123456789']
11 ['123456789']
12 ['123456789']
13 ['123456789']
14 ['123456789']
15 ['123456789']
16 ['123456789']
17 ['123456789']
18 ['123456789']
19 ['123456789']

The kiddy way

def wrap(string, max_width):
    i=0
    strings = []
    s = ""
    for x in string:
        i+=1
        if i == max_width:
            s = s + x
            strings.append(s)
            s = ""
            i = 0
        else:
            s = s + x
    strings.append(s)
    return strings

wrap('ABCDEFGHIJKLIMNOQRSTUVWXYZ',4)
# output: ['ABCD', 'EFGH', 'IJKL', 'IMNO', 'QRST', 'UVWX', 'YZ']