Decorating Hex function to pad zeros
118,686
Solution 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'
Solution 2
How about this:
print '0x%04x' % 42
Solution 3
"{:02x}".format(7) # '07'
"{:02x}".format(27) # '1b'
Where
-
:
is the start of the formatting specification for the first argument{}
to.format()
-
02
means "pad the input from the left with0
s to length2
" -
x
means "format as hex with lowercase letters"
You can also do this with f-strings:
f"{7:02x}" # '07'
f"{27:02x}" # '1b'
Solution 4
If just for leading zeros, you can try zfill
function.
'0x' + hex(42)[2:].zfill(4) #'0x002a'
Solution 5
Use *
to pass width and X
for uppercase
print '0x%0*X' % (4,42) # '0x002A'
As suggested by georg and Ashwini Chaudhary
Author by
jon
Updated on July 08, 2022Comments
-
jon almost 2 years
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?