Python 3.5 - Convert bytes object to 16bit hex string (b'\x07\x89' -> '0x0789')

Question

some device returns data (hex bytes) in the form of

data = b'\x07\x89\x00\x00\x12\x34'

How can I convert this to something in the form of the following?

['0x0789', '0x0000', '0x1234']

I already tried compositions of hexlify. I am using Python 3.5.

Answer 1

Take groups of two from your bytes object. Multiply the first value from each group with 16**2. Add the two values. Use hex on the result to convert it to its string representation.

>>> [hex(data[i]*16**2 + data[i+1]) for i in range(0,len(data),2)]
['0x789', '0x0', '0x1234']

I assume that you don't need your strings padded with useless zeros for now.

Answer 2

Use the struct module; it has unpack function which allows to specify the chunk size size (byte, 2-byte, 4-bytes) and endianess in the data. If what you have is big-endian half-word sized data chunks, then the right format key is ">H".

To parse all data at one, add count in the format specifier: for example ">3H" for you input array. You can also write the number of fields dynamically.

Full example:

import struct
data = b'\x07\x89\x00\x00\x12\x34'
d = struct.unpack(">{}H".format(len(data) // 2), data) # extract fields
result = [hex(x) for x in d] # convert to strings

Answer 3

There are two steps:

1. Chunk the input bytestring into 2-byte sequences
2. Display the sequences as hex literals.

You could use array module from stdlib :

import sys
from array import array

a = array('H', data)
if sys.byteorder != 'big':
    a.byteswap() # use big-endian order
result = ['0x%04X' % i for i in a]
# -> ['0x0789', '0x0000', '0x1234']

It is efficient, especially if you need to read data from a file .