Please try this code

import binascii
mytext='☺00AB12☻0200D60000040001♥t'
print binascii.hexlify(mytext)

I've received this output

3f3030414231323f303230304436303030303034303030313f74

Use str.replace to replace that character with the hex representation of its ordinal value. And to get the actual string back you can decode it using string-decode.

>>> s = 'banana'
>>> print s.replace('a', '\\x' + format(ord('a'), 'x'))
b\x61n\x61n\x61
>>> print s.replace('a', '\\x' + format(ord('a'), 'x')).decode('string-escape')
banana

Some more clever google-fu uncovered this code snippet, which gives me the results I expect:

def convertMac(octet):
    mac = [binascii.b2a_hex(x) for x in list(octet)]
    return ":".join(mac)

So "decoding" is really a misnomer, because this code also yields a proper result (with encode and decode appearing to be syntactic sugar for binascii.a2b() and binascii.b2a()):

Python 2.7.3 (default, Apr 10 2013, 06:20:15)
[GCC 4.6.3] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> string = '\x00$8\xa1\x1c2'
>>> string.encode("hex")
'002438a11c32'
>>>

You can transform your string to an integer generator. Apply hexadecimal formatting for each element and intercalate with a separator:

>>> s = "Hello, World!"
>>> ":".join("{:02x}".format(ord(c)) for c in s)
'48:65:6c:6c:6f:2c:20:57:6f:72:6c:64:21

Try this:

hex_str = "0xAD4"
hex_int = int(hex_str, 16)
new_int = hex_int + 0x200
print hex(new_int)

If you don't like the 0x in the beginning, replace the last line with

print hex(new_int)[2:]

The string '1234' is already equivalent to '\x31\x32\x33\x34':

>>> '\x31\x32\x33\x34'
'1234'
>>> '\x31\x32\x33\x34' == '1234'
True

thus encoding that to hex then decoding it again is.. busy work:

>>> '1234'.encode('hex').decode('hex')
'1234'

\xhh is just a notation to help you create the values; when echoing a byte string Python will always display printable ASCII characters directly rather than use the \xhh notation.

Hex notation here is just a way to express the values of each byte, which is really an integer between 0 and 255. Each byte in a Python string is then a byte with such a constrained value, and encoding to the 'hex' codec produces a string with hex digits for those bytes, and bytes again from the hex digits.

As such, all you have to do is add the \x00 null bytes and the length:

MAGICSTRING = '1234'
value = '\x00{}\x00{}'.format(MAGICSTRING, chr(len(MAGICSTRING) + 2))

Here the \xhh notation is used to produce the null bytes, and the chr() function produces the length 'byte'.

Demo:

>>> MAGICSTRING = '1234'
>>> '\x00{}\x00{}'.format(MAGICSTRING, chr(len(MAGICSTRING) + 2))
'\x001234\x00\x06'
>>> '\x00{}\x00{}'.format(MAGICSTRING, chr(len(MAGICSTRING) + 2)) == '\x00\x31\x32\x33\x34\x00\x06'
True

Given that your string contains ascii only (each char is in range 0-0xff), you can use the following snippet:

In [28]: s = '746865'

In [29]: import math

In [30]: int(s, base=16).to_bytes(math.ceil(len(s) / 2), byteorder='big').decode('ascii')
Out[30]: 'the'

Firstly you need to convert a string into integer with base of 16, then convert it to bytes (assuming 2 chars per byte) and then convert bytes back to string using decode

The hex() function converts a specified integer number into a hexadecimal string representation.

Use int(x, base) with 16 as base to convert the string x to an integer. Call hex(number) with the integer as number to convert it to hexadecimal.

hex_string = "0xAA"

"0x" also required

an_integer = int(hex_string, 16)

hex_value = hex(an_integer)

print(hex_value)

Output

0xaa

Your class.function expects an integer which can be represented either by a decimal or a hexadecimal literal, so that these two calls are completely equivalent:

class.function(0x77)
class.function(119)  # 0x77 == 119

Even print(0x77) will show 119 (because decimal is the default representation).

So, we should rather be talking about converting a string representation to integer. The string can be a hexadecimal representation, like '0x77', then parse it with the base parameter:

 >>> int('0x77', 16)
 119

or a decimal one, then parse it as int('119').

Still, storing integer whenever you deal with integers is better.

EDIT: as @gnibbler suggested, you can parse as int(x, 0), which handles both formats.