12

I wanted to convert an object of type bytes to binary representation in python 3.x.

For example, I want to convert the bytes object b'\x11' to the binary representation 00010001 in binary (or 17 in decimal).

I tried this:

print(struct.unpack("h","\x11"))

But I'm getting:

error struct.error: unpack requires a bytes object of length 2
kiyah
  • 1,488
  • 1
  • 19
  • 26
KaramJaber
  • 777
  • 5
  • 11
  • 22

2 Answers2

16

Starting from Python 3.2, you can use int.from_bytes.

Second argument, byteorder, specifies endianness of your bytestring. It can be either 'big' or 'little'. You can also use sys.byteorder to get your host machine's native byteorder.

import sys
int.from_bytes(b'\x11', byteorder=sys.byteorder)  # => 17
bin(int.from_bytes(b'\x11', byteorder=sys.byteorder))  # => '0b10001'
Oleksii Filonenko
  • 1,361
  • 1
  • 16
  • 25
  • I tried finding the bit representation of a UUID using this method. When i used len function on that bit string, it came out to be 127. But UUID's are 128 bit numbers. Can you please explain why the length calculated was 127? – r4v1 Nov 11 '18 at 08:59
  • 0-127 is 128 numbers – MacItaly Jun 12 '20 at 16:28
  • if the bytes has leading zeros, this method does not work. – Jing He Jan 23 '22 at 18:46
7

Iterating over a bytes object gives you 8 bit ints which you can easily format to output in binary representation:

import numpy as np

>>> my_bytes = np.random.bytes(10)
>>> my_bytes
b'_\xd9\xe97\xed\x06\xa82\xe7\xbf'

>>> type(my_bytes)
bytes

>>> my_bytes[0]
95

>>> type(my_bytes[0])
int

>>> for my_byte in my_bytes:
>>>     print(f'{my_byte:0>8b}', end=' ')

01011111 11011001 11101001 00110111 11101101 00000110 10101000 00110010 11100111 10111111

A function for a hex string representation is builtin:

>>> my_bytes.hex(sep=' ')
'5f d9 e9 37 ed 06 a8 32 e7 bf'
Jim
  • 1,089
  • 9
  • 15