How do I convert a hex triplet to an RGB tuple and back?

Question

I\'d like to convert a hex triplet to an RGB tuple and then convert a tuple to a hex triplet.

Answer 1

A very simplistic approach to convert rgb to hex

>>> rgb = (255, 255, 255)
>>> r, g , b = rgb
>>> hex(r)
'0xff'
>>> hex(r) + hex(g)[2:] + hex(b)[2:]
'0xffffff'
>>>

A simplistic approach to convert Hex to rgb

>>> h  = '0xffffff'
>>> h1, h2, h3 = h[0:4], '0x' + h[4:6], '0x' + h[6:8]
>>> h1, h2, h3
('0xff', '0xff', '0xff')
>>> r, g , b = int(h1, 16), int(h2, 16), int(h3, 16)
>>> r, g, b
(255, 255, 255)

Use a module which provides some these facility: webcolors

>>> hex_to_rgb('#000080')
(0, 0, 128)
>>> rgb_to_hex((255, 255, 255))
'#ffffff'

Function doc:

hex_to_rgb(hex_value) Convert a hexadecimal color value to a 3-tuple of integers suitable for use in an rgb() triplet specifying that color.

rgb_to_hex(rgb_triplet) : Convert a 3-tuple of integers, suitable for use in an rgb() color triplet, to a normalized hexadecimal value for that color.

Answer 2

HEX to RGB tuple

>>> tuple(bytes.fromhex('61559a'))
(97, 85, 154)

RGB tuple to HEX

>>> bytes((97, 85, 154)).hex()
'61559a'

No imports needed!

---

What is this magic?!

Since bytes objects are sequences of integers (akin to a tuple), for a bytes object b, b[0] will be an integer, while b[0:1] will be a bytes object of length 1

...

The representation of bytes objects uses the literal format (b'...') since it is often more useful than e.g. bytes([46, 46, 46]). You can always convert a bytes object into a list of integers using list(b).

Source: https://docs.python.org/3/library/stdtypes.html#bytes-objects

Answer 3

You can use a look-up table with some slicing and shifts — all relatively fast operations — to create a couple of functions that will work unchanged in both Python 2 and 3:

_NUMERALS = '0123456789abcdefABCDEF'
_HEXDEC = {v: int(v, 16) for v in (x+y for x in _NUMERALS for y in _NUMERALS)}
LOWERCASE, UPPERCASE = 'x', 'X'

def rgb(triplet):
    return _HEXDEC[triplet[0:2]], _HEXDEC[triplet[2:4]], _HEXDEC[triplet[4:6]]

def triplet(rgb, lettercase=LOWERCASE):
    return format(rgb[0]<<16 | rgb[1]<<8 | rgb[2], '06'+lettercase)

if __name__ == '__main__':
    print('{}, {}'.format(rgb('aabbcc'), rgb('AABBCC')))
    # -> (170, 187, 204), (170, 187, 204)

    print('{}, {}'.format(triplet((170, 187, 204)),
                          triplet((170, 187, 204), UPPERCASE)))
    # -> aabbcc, AABBCC

    print('{}, {}'.format(rgb('aa0200'), rgb('AA0200')))
    # -> (170, 2, 0), (170, 2, 0)

    print('{}, {}'.format(triplet((170, 2, 0)),
                          triplet((170, 2, 0), UPPERCASE)))
    # -> aa0200, AA0200

Answer 4

>>> import struct
>>> rgbstr='aabbcc'
>>> struct.unpack('BBB',rgbstr.decode('hex'))
(170, 187, 204)

and

>>> rgb = (50,100,150)
>>> struct.pack('BBB',*rgb).encode('hex')
'326496'

Answer 5

Trying to be pythonic:

>>> rgbstr='aabbcc'
>>> tuple(ord(c) for c in rgbstr.decode('hex'))
(170, 187, 204)
>>> tuple(map(ord, rgbstr.decode('hex'))
(170, 187, 204)

and

>>> rgb=(12,50,100)
>>> "".join(map(chr, rgb)).encode('hex')
'0c3264'

Answer 6

import re

def hex_to_int_color(v):
  return tuple(int(i,16) for i in re.match(
    r'^#?([0-9a-f]{2})([0-9a-f]{2})([0-9a-f]{2})$', v,
    flags=re.IGNORECASE).groups())

def int_to_hex_color(v):
  return '#%02x%02x%02x' % v