RGB Int to RGB - Python

Question

How can I convert an RGB integer to the corresponding RGB tuple (R,G,B)? Seems simple enough, but I can't find anything on google.

I know that for every RGB (r,g,b) you have the integer n = r256^2 + g256 + b, how can I solve the reverse in Python, IE given an n, I need the r,g,b values.

Answer 1

I'm not a Python expert by all means, but as far as I know it has the same operators as C.

If so this should work and it should also be a lot quicker than using modulo and division.

Blue =  RGBint & 255
Green = (RGBint >> 8) & 255
Red =   (RGBint >> 16) & 255

What it does it to mask out the lowest byte in each case (the binary and with 255.. Equals to a 8 one bits). For the green and red component it does the same, but shifts the color-channel into the lowest byte first.

Answer 2

From a RGB integer:

Blue =  RGBint mod 256
Green = RGBint / 256 mod 256
Red =   RGBint / 256 / 256 mod 256

This can be pretty simply implemented once you know how to get it. :)

Upd: Added python function. Not sure if there's a better way to do it, but this works on Python 3 and 2.4

def rgb_int2tuple(rgbint):
    return (rgbint // 256 // 256 % 256, rgbint // 256 % 256, rgbint % 256)

There's also an excellent solution that uses bitshifting and masking that's no doubt much faster that Nils Pipenbrinck posted.

Answer 3

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

for python3:
>>> struct.unpack('BBB', bytes.fromhex(str))

and

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

for python3:
>>> bytes.hex(struct.pack('BBB',*rgb))

Answer 4

>>> r, g, b = (111, 121, 131)
>>> packed = int('%02x%02x%02x' % (r, g, b), 16)

This produces the following integer:

>>> packed
7305603

You can then unpack it either the long explicit way:

>>> packed % 256
255
>>> (packed / 256) % 256
131
>>> (packed / 256 / 256) % 256
121
>>> (packed / 256 / 256 / 256) % 256
111

..or in a more compact manner:

>>> b, g, r = [(packed >> (8*i)) & 255 for i in range(3)]
>>> r, g, b

Sample applies with any number of digits, e.g an RGBA colour:

>>> packed = int('%02x%02x%02x%02x' % (111, 121, 131, 141), 16)
>>> [(packed >> (8*i)) & 255 for i in range(4)]
[141, 131, 121, 111]

Answer 5

I assume you have a 32-bit integer containing the RGB values (e.g. ARGB). Then you can unpack the binary data using the struct module:

# Create an example value (this represents your 32-bit input integer in this example).
# The following line results in exampleRgbValue = binary 0x00FF77F0 (big endian)
exampleRgbValue = struct.pack(">I", 0x00FF77F0)

# Unpack the value (result is: a = 0, r = 255, g = 119, b = 240)
a, r, g, b = struct.unpack("BBBB", exampleRgbValue)

Answer 6

Just a note for anyone using Google's Appengine Images Python API. I found I had a situation where I had to supply a method with a 32-bit RGB color value.

Specifically, if you're using the API to convert a PNG (or any image with transparent pixels), you'll need to supply the execute_transforms method with an argument called transparent_substitution_rgb which has to be a 32-bit RGB color value.

Borrowing from dbr's answer, I came up with a method similar to this:

def RGBTo32bitInt(r, g, b):
  return int('%02x%02x%02x' % (r, g, b), 16)

transformed_image = image.execute_transforms(output_encoding=images.JPEG, transparent_substitution_rgb=RGBTo32bitInt(255, 127, 0))

Answer 7

def unpack2rgb(intcol):
    tmp, blue= divmod(intcol, 256)
    tmp, green= divmod(tmp, 256)
    alpha, red= divmod(tmp, 256)
    return alpha, red, green, blue

If only the divmod(value, (divider1, divider2, divider3…)) suggestion was accepted, it would have simplified various time conversions too.

Answer 8

There's probably a shorter way of doing this:

dec=10490586
hex="%06x" % dec
r=hex[:2]
g=hex[2:4]
b=hex[4:6]
rgb=(r,g,b)

EDIT: this is wrong - gives the answer in Hex, OP wanted int. EDIT2: refined to reduce misery and failure - needed '%06x' to ensure hex is always shown as six digits [thanks to Peter Hansen's comment].

Answer 9

If you are using NumPy and you have an array of RGBints, you can also just change its dtype to extract the red, green, blue and alpha components:

>>> type(rgbints)
numpy.ndarray
>>> rgbints.shape
(1024L, 768L)
>>> rgbints.dtype
dtype('int32')
>>> rgbints.dtype = dtype('4uint8')
>>> rgbints.shape
(1024L, 768L, 4L)
>>> rgbints.dtype
dtype('uint8')

Answer 10

def int2color(red,green,blue):
    def chex(c):
        if c < 16:
            return "0"+hex(c)[2:]
        else:
            return hex(c)[2:]

    return "#{}{}{}".format(chex(red),chex(green),chex(blue))

>>> int2color(49, 105, 0)
#316900

Answer 11

Adding to what is mentioned above. A concise one-liner alternative.

# 2003199 or #E190FF is Dodger Blue.
tuple((2003199 >> Val) & 255 for Val in (16, 8, 0))
# (30, 144, 255)

And to avoid any confusion in the future.

from collections import namedtuple
RGB = namedtuple('RGB', ('Red', 'Green', 'Blue'))
rgb_integer = 16766720  # Or #ffd700 is Gold.
# The unpacking asterisk prevents a TypeError caused by missing arguments.
RGB(*((rgb_integer >> Val) & 255 for Val in (16, 8, 0)))
# RGB(Red=255, Green=215, Blue=0)