[python] Python 2,3 Convert Integer to "bytes" Cleanly

The shortest ways I have found are:

n = 5

# Python 2.
s = str(n)
i = int(s)

# Python 3.
s = bytes(str(n), "ascii")
i = int(s)

I am particularly concerned with two factors: readability and portability. The second method, for Python 3, is ugly. However, I think it may be backwards compatible.

Is there a shorter, cleaner way that I have missed? I currently make a lambda expression to fix it with a new function, but maybe that's unnecessary.

from int to byte:

bytes_string = int_v.to_bytes( lenth, endian )

where the lenth is 1/2/3/4...., and endian could be 'big' or 'little'

form bytes to int:

data_list = list( bytes );

I have found the only reliable, portable method to be

bytes(bytearray([n]))

Just bytes([n]) does not work in python 2. Taking the scenic route through bytearray seems like the only reasonable solution.

You can use the struct's pack:

In [11]: struct.pack(">I", 1)
Out[11]: '\x00\x00\x00\x01'

The ">" is the byte-order (big-endian) and the "I" is the format character. So you can be specific if you want to do something else:

In [12]: struct.pack("<H", 1)
Out[12]: '\x01\x00'

In [13]: struct.pack("B", 1)
Out[13]: '\x01'

This works the same on both python 2 and python 3.

Note: the inverse operation (bytes to int) can be done with unpack.

In Python 3.x, you can convert an integer value (including large ones, which the other answers don't allow for) into a series of bytes like this:

import math

x = 0x1234
number_of_bytes = int(math.ceil(x.bit_length() / 8))

x_bytes = x.to_bytes(number_of_bytes, byteorder='big')

x_int = int.from_bytes(x_bytes, byteorder='big')
x == x_int

Converting an int to a byte in Python 3:

    n = 5    
    bytes( [n] )

>>> b'\x05'

;) guess that'll be better than messing around with strings

source: http://docs.python.org/3/library/stdtypes.html#binaryseq