Convert from 8 bits array to 5 bits array

Question

Is there easy way for python to convert from 8 bits to 5 bits. Currently I am using this code to do it:

def convertbits(data, frombits, tobits, pad=True):
    acc = 0
    bits = 0
    ret = []
    maxv = (1 << tobits) - 1
    max_acc = (1 << (frombits + tobits - 1)) - 1
    for value in data:
        if value < 0 or (value >> frombits):
            return None
        acc = ((acc << frombits) | value) & max_acc
        bits += frombits
        while bits >= tobits:
            bits -= tobits
            ret.append((acc >> bits) & maxv)
    if pad:
        if bits:
            ret.append((acc << (tobits - bits)) & maxv)
    elif bits >= frombits or ((acc << (tobits - bits)) & maxv):
        return None
    return ret

is there a better way?

Edit: output should be list of 5bit integers without loosing any data in proccess

it should work like:

>>> hello = [ord(letter) for letter in 'hello']
>>> hello
[104, 101, 108, 108, 111]
>>> convertbits(hello, 8, 5)
[13, 1, 18, 22, 24, 27, 3, 15]
>>> 

Answer 1

Well, this is relatively memory-inefficient as it converts individual bits to strings, but it seems to work:

import itertools  

def convertbits(data, From, To):
    bits_orig = itertools.chain.from_iterable(bin(n)[2:].zfill(From) for n in data)

    chunks = iter(lambda: ''.join(itertools.islice(bits_orig, To)), '')

    return [int(n, 2) for n in chunks]


print(convertbits(b'hello', 8, 5))
print([13, 1, 18, 22, 24, 27, 3, 15])

Once you got a stream of bits of the numbers ( bits_orig ), it's then simple to slice this stream into chunks of equal length ( chunks ) (this version doesn't do padding but it's fairly simple to implement) and convert the strings of ones and zeros back to numbers.

---

If you're working with 8-bit numbers exclusively, here's an algorithm that's 8.5 (!) times faster than the one above:

from collections import deque 

def convert8bits(data, To):
    number = int.from_bytes(data, 'big')

    ret = deque()
    th = (1 << To) - 1
    while number:
        ret.appendleft(number & th)
        number >>= To

    return ret