Issue
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
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'
Answered By - Tim Pietzcker
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