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

Starting with Python 3.6, you can:

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

---

for older python versions use the .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'