[python] Format ints into string of hex

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.

Example:

Input: [0,1,2,3,127,200,255], 
Output: 000102037fc8ff

I've managed to come up with:

#!/usr/bin/env python

def format_me(nums):
    result = ""
    for i in nums:
        if i <= 9:
            result += "0%x" % i
        else:
            result += "%x" % i
    return result

print format_me([0,1,2,3,127,200,255])

However, this looks a bit awkward. Is there a simpler way?

Example with some beautifying, similar to the sep option available in python 3.8

def prettyhex(nums, sep=''):
    return sep.join(f'{a:02x}' for a in nums)

numbers = [0, 1, 2, 3, 127, 200, 255]
print(prettyhex(numbers,'-'))

output

00-01-02-03-7f-c8-ff

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'.

With python 2.X, you can do the following:

numbers = [0, 1, 2, 3, 127, 200, 255]
print "".join(chr(i).encode('hex') for i in numbers)

print

'000102037fc8ff'

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

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

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

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

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.)

Yet another option is binascii.hexlify:

a = [0,1,2,3,127,200,255]
print binascii.hexlify(bytes(bytearray(a)))

prints

000102037fc8ff

This is also the fastest version for large strings on my machine.

In Python 2.7 or above, you could improve this even more by using

binascii.hexlify(memoryview(bytearray(a)))

saving the copy created by the bytes call.

Python 2:

>>> str(bytearray([0,1,2,3,127,200,255])).encode('hex')
'000102037fc8ff'

Python 3:

>>> bytearray([0,1,2,3,127,200,255]).hex()
'000102037fc8ff'

From Python documentation. Using the built in format() function you can specify hexadecimal base using an 'x' or 'X' Example:

x= 255 print('the number is {:x}'.format(x))

Output:

the number is ff

Here are the base options

Type
'b' Binary format. Outputs the number in base 2. 'c' Character. Converts the integer to the corresponding unicode character before printing. 'd' Decimal Integer. Outputs the number in base 10. 'o' Octal format. Outputs the number in base 8. 'x' Hex format. Outputs the number in base 16, using lower- case letters for the digits above 9. 'X' Hex format. Outputs the number in base 16, using upper- case letters for the digits above 9. 'n' Number. This is the same as 'd', except that it uses the current locale setting to insert the appropriate number separator characters. None The same as 'd'.