How to convert hexadecimal string to character with that code point?

Question

I have the string x = '0x32' and would like to turn it into y = '\x32'.
Note that len(x) == 4 and len(y) == 1.

I've tried to use z = x.replace("0", "\\"), but that causes z = '\\x32' and len(z) == 4. How can I achieve this?

Answer 1

You do not have to make it that hard: you can use int(..,16) to parse a hex string of the form 0x.... Next you simply use chr(..) to convert that number into a character with that Unicode (and in case the code is less than 128 ASCII) code:

y = chr(int(x,16))

This results in:

>>> chr(int(x,16))
'2'

But \x32 is equal to '2' (you can look it up in the ASCII table):

>>> chr(int(x,16)) == '\x32'
True

and:

>>> len(chr(int(x,16)))
1

Answer 2

Try this:

z = x[2:].decode('hex')

Answer 3

The ability to include code points like '\x32' inside a quoted string is a convenience for the programmer that only works in literal values inside the source code. Once you're manipulating strings in memory, that option is no longer available to you, but there are other ways of getting a character into a string based on its code point value.

Also note that '\x32' results in exactly the same string as '2'; it's just typed out differently.

Given a string containing a hexadecimal literal, you can convert it to its numeric value with int(str,16). Once you have a numeric value, you can convert it to the character with that code point via chr(). So putting it all together:

x = '0x32'
print(chr(int(x,16)))
#=> 2