Cut a binary number into groups of five digits

Question

Is there any neat trick to chop a binary number into groups of five digits in python?

'00010100011011101101110100010111' => ['00010', '00110', '10111', ...]

Edit: I want to write a cipher / encoder to create "easy to read by phone" tokens. Base32 standard encoding has the following disadvantages:

• The potential for generating random words f *
• Uses confused characters, such as the characters "I", "L", "O" (can be confused with 0 and 1)
• Easy to guess sequences ("AAAA", "AAAB", ...)

I was able to knock over myself in 20 lines of python, thanks to everyone. My encoder leaves "I", "L", "O" and "U", and the resulting sequences are hard to guess.

Answer 1

>>> a='00010100011011101101110100010111'
>>> [a[i:i+5] for i in range(0, len(a), 5)]
['00010', '10001', '10111', '01101', '11010', '00101', '11']

Answer 2

>>> [''.join(each) for each in zip(*[iter(s)]*5)]
['00010', '10001', '10111', '01101', '11010', '00101']

or

>>> map(''.join, zip(*[iter(s)]*5))
['00010', '10001', '10111', '01101', '11010', '00101']

[EDIT]

The question was raised by Greg Hugill, what to do with two tails? Here are a few possibilities:

>>> from itertools import izip_longest
>>>
>>> map(''.join, izip_longest(*[iter(s)]*5, fillvalue=''))
['00010', '10001', '10111', '01101', '11010', '00101', '11']
>>>
>>> map(''.join, izip_longest(*[iter(s)]*5, fillvalue=' '))
['00010', '10001', '10111', '01101', '11010', '00101', '11   ']
>>>
>>> map(''.join, izip_longest(*[iter(s)]*5, fillvalue='0'))
['00010', '10001', '10111', '01101', '11010', '00101', '11000']

Answer 3

32.

>>> import base64
>>> base64.b32encode("good stuff")
'M5XW6ZBAON2HKZTG'

Answer 4

?

>>> import re
>>> re.findall('.{1,5}', '00010100011011101101110100010111')
['00010', '10001', '10111', '01101', '11010', '00101', '11']

, , .

Answer 5

, .

, Generators

from itertools import islice
def slice_generator(an_iter, num):
    an_iter = iter(an_iter)
    while True:
        result = tuple(islice(an_iter, num))
        if not result:
           return
        yield result

, :

>>> l = '00010100011011101101110100010111'
>>> [''.join(x) for x in slice_generator(l,5)]
['00010', '10001', '10111', '01101', '11010', '00101', '11']

Answer 6

>>> l = '00010100011011101101110100010111'
>>> def splitSize(s, size):
...     return [''.join(x) for x in zip(*[list(s[t::size]) for t in range(size)])]
...  
>>> splitSize(l, 5)
['00010', '10001', '10111', '01101', '11010', '00101']
>>> 

Answer 7

Another way to group iterations from itertools examples :

def grouper(n, iterable, fillvalue=None):
    "grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)