int given base 2 and then hex:

>>> int('010110', 2)
22
>>> hex(int('010110', 2))
'0x16'
>>> 

>>> hex(int('0000010010001101', 2))
'0x48d'

The doc of int:

int(x[, base]) -> integer

Convert a string or number to an integer, if possible.  A floating

point argument will be truncated towards zero (this does not include a string representation of a floating point number!) When converting a string, use the optional base. It is an error to supply a base when converting a non-string. If base is zero, the proper base is guessed based on the string content. If the argument is outside the integer range a long object will be returned instead.

The doc of hex:

hex(number) -> string

Return the hexadecimal representation of an integer or long

integer.

For solving the left-side trailing zero problem:

my_hexdata = "1a"

scale = 16 ## equals to hexadecimal

num_of_bits = 8

bin(int(my_hexdata, scale))[2:].zfill(num_of_bits)

It will give 00011010 instead of the trimmed version.

You can use ord and hex like this :

>>> s = 'some string'
>>> hex_chars = map(hex,map(ord,s))
>>> print hex_chars
['0x73', '0x6f', '0x6d', '0x65', '0x20', '0x73', '0x74', '0x72', '0x69', '0x6e', '0x67']
>>> hex_string = "".join(c[2:4] for c in hex_chars)
>>> print hex_string
736f6d6520737472696e67
>>>

Or use the builtin encoding :

>>> s = 'some string'
>>> print s.encode('hex_codec')
736f6d6520737472696e67
>>>

To format to hexadecimal you can use the hex function:

>>> hex(int('10000011101000011010100010010111', 2))
0x83a1a897

Or to get it in exactly the format you requested:

>>> '%08X' % int('10000011101000011010100010010111', 2)
83A1A897

import base64
data = base64.b16decode("01000001B8000102030405")