Python: convert a byte to binary and shift its bits?

Question

I want to convert an encoding to another encoding in Python for a school project. However, the encoding I am translating from will add a padding to its encoding on the first bit.

How do I shift an binary number sequence to the left by one, so that it goes from:

00000001 11001100 01010101 and so on

to

00000011 10011000 10101010 and so on

so the end result's lowest bit will be the former's highest bit number?

how many bytes does it have? is it 3 or was that just an example? — Dleep, Aug 07 '15 at 18:53
At some point you'll need to test the leftmost bit before doing a left shift. You can optimize how often you'll need to do that by eg. packing multiple bytes into a larger int (check the `struct` module). — thebjorn, Aug 07 '15 at 18:54
just 3 is an example, but its a continuously growing number. As its actually data transmitted from somewhere and its needs to be converted out to be printed in a different format. — needoriginalname, Aug 07 '15 at 18:58

score 5 · Answer 1 · answered Aug 07 '15 at 18:54

5

You can use the << operator to left shift, and conversely >> will right shift

>>> x = 7485254
>>> bin(x)
'0b11100100011011101000110'
>>> bin(x << 1)
'0b111001000110111010001100'

answered Aug 07 '15 at 18:54

Cory Kramer

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score 1 · Answer 2 · answered May 12 '16 at 09:43

You could use the bitstring library which allows for bitwise operations on arbitrarily long bitstrings e.g. to import and shift your binary number:

>>> import bitstring
>>> bitstring.BitArray(bin='0b11100100011011101000110') << 1
BitArray('0b11001000110111010001100')

score -3 · Accepted Answer · answered Aug 07 '15 at 19:06

-3

You can convert the string to one big integer and then do the left-shift (and then convert the big integer back into a string):

large_int = bytes2int(mystring)
large_int <<= 1
mystring = int2bytes(large_int)

using e.g. this simplistic implementation:

def bytes2int(str):
    res = ord(str[0])
    for ch in str[1:]:
        res <<= 8
        res |= ord(ch)
    return res

def int2bytes(n):
    res = []
    while n:
        ch = n & 0b11111111
        res.append(chr(ch))
        n >>= 8
    return ''.join(reversed(res))

bytes = 'abcdefghijklmnopqrstuv'
assert int2bytes(bytes2int(bytes)) == bytes

answered Aug 07 '15 at 19:06

thebjorn

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It is inefficient to process one byte at a time. You could use `int.from_bytes()`, `int.to_bytes()` methods on Python 3 and [`binascii.unhexlify` on Python 2](http://stackoverflow.com/a/28524760/4279). See also, [Convert Binary to ASCII and vice versa (Python)](http://stackoverflow.com/q/7396849/4279) – jfs Aug 08 '15 at 22:53
@J.F.Sebastian you're absolutely correct, hence the note about 'simplistic' implementation (the OP is doing binary encoding work, so I figured a bit-fiddling example could be interesting). – thebjorn Aug 09 '15 at 13:44
If you want to show byte at a time operation; you could [use `bytearray()` that is Python 2/3, Pypy compatible](http://stackoverflow.com/a/15916760/4279) – jfs Aug 09 '15 at 13:56
1

Is anyone actually using Python 3/Pypy? – thebjorn Aug 09 '15 at 15:33
I just tested `bytearray` performance on Python 2.7.10 (using the example at the bottom of http://www.dotnetperls.com/bytes), and it seems like it is almost 10% slower than a plain list. FWIW, I would expect PyPy to be able to optimize the functions in the answer without any problems (Cython makes it about twice as fast with very few changes). – thebjorn Aug 09 '15 at 18:17

Python: convert a byte to binary and shift its bits?

3 Answers3