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 o         


        

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'

Answer 2

Suppose you want to have leading zeros for hexadecimal number, for example you want 7 digit where your hexadecimal number should be written on, you can do like that :

hexnum = 0xfff
str_hex =  hex(hexnum).rstrip("L").lstrip("0x") or "0"
'0'* (7 - len(str_hexnum)) + str_hexnum

This gives as a result :

'0000fff'

Answer 3

If just for leading zeros, you can try zfill function.

'0x' + hex(42)[2:].zfill(4) #'0x002a'

Answer 4

How about this:

print '0x%04x' % 42

Answer 5

What I needed was

"{:02x}".format(2)   # '02'
"{:02x}".format(42)  # '2a'

or as f-strings:

f"{2:02x}"   # '02'
f"{42:02x}"  # '2a'

Answer 6

Use * to pass width and X for uppercase

print '0x%0*X' % (4,42) # '0x002A'

As suggested by georg and Ashwini Chaudhary