Format ints into string of hex

Question

I need to create a string of hex digits from a list of random integers (0-255). Each hex digit should be represented by two characters: 5 - \"05\", 16 - \"10\", etc.

Answer 1

Just for completeness, using the modern .format() syntax:

>>> numbers = [1, 15, 255]
>>> ''.join('{:02X}'.format(a) for a in numbers)
'010FFF'

Answer 2

Similar to my other answer, except repeating the format string:

>>> numbers = [1, 15, 255]
>>> fmt = '{:02X}' * len(numbers)
>>> fmt.format(*numbers)
'010FFF'

Answer 3

The most recent and in my opinion preferred approach is the f-string:

''.join(f'{i:02x}' for i in [1, 15, 255])

Format options

The old format style was the %-syntax:

['%02x'%i for i in [1, 15, 255]]

The more modern approach is the .format method:

 ['{:02x}'.format(i) for i in [1, 15, 255]]

More recently, from python 3.6 upwards we were treated to the f-string syntax:

[f'{i:02x}' for i in [1, 15, 255]]

Format syntax

Note that the f'{i:02x}' works as follows.

• The first part before : is the input or variable to format.
• The x indicates that the string should be hex. f'{100:02x}' is '64' and f'{100:02d}' is '1001'.
• The 02 indicates that the string should be left-filled with 0's to length 2. f'{100:02x}' is '64' and f'{100:30x}' is ' 64'.

Answer 4

a = [0,1,2,3,127,200,255]
print str.join("", ("%02x" % i for i in a))

prints

000102037fc8ff

(Also note that your code will fail for integers in the range from 10 to 15.)