Decorating Hex function to pad zeros

Question

I wrote this simple function:

def padded_hex(i, l):     given_int = i     given_len = l      hex_result = hex(given_int)[2:] # remove '0x' from beginning of str     num_hex_chars = len(hex_result)     extra_zeros = '0' * (given_len - num_hex_chars) # may not get used..      return ('0x' + hex_result if num_hex_chars == given_len else             '?' * given_len if num_hex_chars > given_len else             '0x' + extra_zeros + hex_result if num_hex_chars < given_len else             None) 

Examples:

padded_hex(42,4) # result '0x002a' hex(15) # result '0xf' padded_hex(15,1) # result '0xf' 

Whilst this is clear enough for me and fits my use case (a simple test tool for a simple printer) I can't help thinking there's a lot of room for improvement and this could be squashed down to something very concise.

What other approaches are there to this problem?

Answer 1

Use the new .format() string method:

>>> "{0:#0{1}x}".format(42,6) '0x002a' 

Explanation:

{   # Format identifier 0:  # first parameter #   # use "0x" prefix 0   # fill with zeroes {1} # to a length of n characters (including 0x), defined by the second parameter x   # hexadecimal number, using lowercase letters for a-f }   # End of format identifier 

If you want the letter hex digits uppercase but the prefix with a lowercase 'x', you'll need a slight workaround:

>>> '0x{0:0{1}X}'.format(42,4) '0x002A' 

Starting with Python 3.6, you can also do this:

>>> value = 42 >>> padding = 6 >>> f"{value:#0{padding}x}" '0x002a'