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pythonhexpaddingbuilt-in

Decorating Hex function to pad zeros


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?


Solution

  • 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'