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
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'
INTEGER.to_bytes(..)
''.join(f'{byte:08b}' for byte in my_bytes)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'
Use bitstring package
import bitstring
bitstring.BitArray(b"\x5a\x01\x00").bin
Returns '010110100000000100000000'
Warning
The accepted answer bin(int.from_bytes(b"\x5a\x01\x00", byteorder=sys.byteorder)) returns '0b101011010' which is obviously nonsense!
The accepted answer might fail if you convert int.from_bytes(b'\x11\x00', byteorder=sys.byteorder). It might be also contraintuitive to represent bytes in little endian int, when it swap the bytes order and I don't know what's going on under the hood when dealing with e.g. sequence of 9 bytes.
