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How to convert a RGBA color tuple, example (96, 96, 96, 202), to corresponding RGB color tuple?

Edit:

What I want is to get a RGB value which is most similar to the RGBA tuple visually on white background.

Stéphane Bruckert
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jack
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    Wouldn't that just be (96,96,96)? – Lazarus Jan 12 '10 at 13:33
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    That depends on the background pixel's colour. – Frank Bollack Jan 12 '10 at 13:34
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    Do you just want to remove the alpha channel? Or are do you want the RGB result of superimposing the RGBA over a (e.g.) white background? – Ben James Jan 12 '10 at 13:35
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    You might want to read the famous paper "Compositing digital images" (by Porter and Duff) for full details on alpha-composition: http://keithp.com/~keithp/porterduff/ – kusma Jan 12 '10 at 15:30
  • The answer of Andras Zoltan and hkurabko are also useful to calculate the oposite, I mean if you have various alpha blended colors and have its original backgrounds (mattes) then you could calculate the original RGBA color which is what I've been looking for a while ;) – nacho4d Jul 17 '10 at 06:54

9 Answers9

101

I've upvoted Johannes' answer because he's right about that.

* A few comments have been raised that my original answer was not correct. It worked if alpha values were inverted from the normal. By definition, however, this won't work in most cases. I've therefore updated the formula below to be correct for the normal case. This ends up being equal to @hkurabko's answer below *

A more specific answer, however, incorporates the alpha value into the actual colour result based on an opaque background colour (or 'matte' as it's referred to).

There is an algorithm for this (from this wikipedia link):

  • Normalise the RGBA values so that they're all between 0 and 1 - just divide each value by 255 to do this. We'll call the result Source.
  • Normalise also the matte colour (black, white whatever). We'll call the result BGColor Note - if the background colour is also transparent, then you'll have to recurse the process for that first (again, choosing a matte) to get the source RGB for this operation.
  • Now, the conversion is defined as (in complete psuedo code here!):

    Source => Target = (BGColor + Source) =
    Target.R = ((1 - Source.A) * BGColor.R) + (Source.A * Source.R)
    Target.G = ((1 - Source.A) * BGColor.G) + (Source.A * Source.G)
    Target.B = ((1 - Source.A) * BGColor.B) + (Source.A * Source.B)
    

To get the final 0-255 values for Target you simply multiply all the normalised values back up by 255, making sure you cap at 255 if any of the combined values exceed 1.0 (this is over-exposure and there are more complex algorithms dealing with this that involve whole-image processing etc.).

EDIT: In your question you said you want a white background - in that case just fix BGColor to 255,255,255.

Andras Zoltan
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    In the light of the question edit this is a very accurate and good answer. I deleted mine and hope you bubble up, since you explained the matters way better :) – Joey Jan 12 '10 at 14:08
  • I think this answer assumes that the background color is rgb, not rgba. – Leo Apr 03 '14 at 02:21
  • This is unusual considering that RGBA(0,0,0,1) is fully opaque black. But with the given algorithm, it would be fully transparent black. I think rearranging the algorithm to reflect this would make it a more helpful resource and better answer to the question. – Bryan Rayner Dec 04 '14 at 18:01
  • Yes, I concede that the alpha channel is often expressed the opposite way around - with 0 being transparent and 1 being opaque. @hkurabko's answer does it the other way around, so rather than change my answer that, I think, is why that answer is gaining upvotes in addition to mine. – Andras Zoltan Dec 04 '14 at 21:39
  • that said - looking more closely to that answer, it appears to be identical to mine. – Andras Zoltan Dec 04 '14 at 21:42
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    @BryanRayner have added some edits to my answer. I'm gaining upvotes for this answer regularly, and it's possible that some people are finding it, having to invert the algorithm themselves. It'd be better if it was more complete. Thanks. – Andras Zoltan Dec 04 '14 at 21:51
  • @Andras Zoltan, there's a difference in your sample code from hkurabko's. It's definitely not quite right when using it just on paper. Unless that's intentional, I think you have an error. – Bryan Rayner Dec 05 '14 at 20:31
  • I have now changed my solution to use alpha the right way around; I needed to use this just now; and having it the wrong way round just doesn't make any sense in the general case! – Andras Zoltan Jan 26 '15 at 08:44
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    If anyone is still interested, this JSBin allows to watch the color being transformed. Also it is ready to be used as a javacript function. I used BGColor as white, because I just do not understand the matte concept (sorry!). The link is as follows: http://jsbin.com/qocixeh/edit?html,js,console,output and as well a gist: https://gist.github.com/vladimirbrasil/bd139851ff757a1c8cb46fac93e733eb Hope it helps too. A zillion thanks for the answer! – Vladimir Brasil Dec 07 '17 at 20:13
  • I have just asked a simular question here https://stackoverflow.com/questions/58574084/convert-argb-hex-string-to-rgb - I want to convert a hex string Argb to Rgb. Your calculation is great but I don't know how to convert from/to hex. – Luke T O'Brien Oct 26 '19 at 19:39
  • Awesome Post! Quick question- How do we have BGColor.R, BGColor.B, BGColor.G if there is only 1 value of opaqueness per pixel? – maximus Dec 09 '21 at 18:07
44

hm... regarding to

http://en.wikipedia.org/wiki/Alpha_compositing#Alpha_blending

solution provided by Andras Zoltan should be slightly changed to:

Source => Target = (BGColor + Source) =
Target.R = ((1 - Source.A) * BGColor.R) + (Source.A * Source.R)
Target.G = ((1 - Source.A) * BGColor.G) + (Source.A * Source.G)
Target.B = ((1 - Source.A) * BGColor.B) + (Source.A * Source.B)

This changed version works fine for me, because in prev. version rgba(0,0,0,0) with matte rgb(ff,ff,ff) will be changed to rgb(0,0,0).

hkurabko
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  • I believe your are right. Wikipedia and Andras Zoltan's answer is a little bit different. – nacho4d Jul 17 '10 at 06:52
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    I agree, use this version, not @Andras Zoltan's. For a test scenario, convert rgba(0,0,0,.7) on background rgb(255,255,255) using both formulas; hkurabko's forumala clearly produces the correct answer. – Ryre Jul 29 '11 at 19:21
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    Andras defines 0 alpha as opaque and 255 as transparent, while CSS defines it the opposite way. This answer works with the CSS's definition of alpha. – Casey Chu Aug 22 '12 at 06:12
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    +1 This works with the most common definition of alpha (0 = transparent) like OpenGL, SDL, graphics editors. – DLight Aug 22 '14 at 14:26
  • +1 - and I have updated my answer to work with alpha being the 'right way round', as per this as well. Ironically, having just tried to use my solution to do this, I agree that my original answer, although it made its own kind of sense, didn't really work for most applications. – Andras Zoltan Jan 26 '15 at 08:45
11

In my case, I wanted to convert an RGBA image to RGB and the following worked just as expected:

rgbImage = cv2.cvtColor(npimage, cv2.COLOR_RGBA2RGB)
illright
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SulabhMatele
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This depends on the color space you use. If the RGBA is in pre-multiplied color-space and is semi-transparent, you need to divide out alpha to get the correct RGB color. If the color is in non pre-multiplied color-space, then you can just discard the alpha channel.

kusma
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3

Here is some java code (works on Android API 24):

        //int rgb_background = Color.parseColor("#ffffff"); //white background
        //int rgba_color = Color.parseColor("#8a000000"); //textViewColor 

        int defaultTextViewColor = textView.getTextColors().getDefaultColor();

        int argb = defaultTextViewColor;
        int alpha = 0xFF & (argb >> 24);
        int red = 0xFF & (argb >> 16);
        int green = 0xFF & (argb >> 8);
        int blue = 0xFF & (argb >> 0);
        float alphaFloat = (float)alpha / 255;

        String colorStr = rgbaToRGB(255, 255, 255, red, green, blue, alphaFloat);

function:

protected String rgbaToRGB(int rgb_background_red, int rgb_background_green, int rgb_background_blue,
                        int rgba_color_red, int rgba_color_green, int rgba_color_blue, float alpha) {

    float red = (1 - alpha) * rgb_background_red + alpha * rgba_color_red;
    float green = (1 - alpha) * rgb_background_green + alpha * rgba_color_green;
    float blue = (1 - alpha) * rgb_background_blue + alpha * rgba_color_blue;

    String redStr = Integer.toHexString((int) red);
    String greenStr = Integer.toHexString((int) green);
    String blueStr = Integer.toHexString((int) blue);

    String colorHex = "#" + redStr + greenStr + blueStr;

    //return Color.parseColor(colorHex);
    return colorHex;
}
live-love
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2

Python function in accordance with hkurabko's answer.

def rgba2rgb(rgba: tuple[int, int, int, float], background: tuple[int, int, int] = (255, 255, 255)):
    return (
        round(((1 - rgba[3]) * background[0]) + (rgba[3] * rgba[0])),
        round(((1 - rgba[3]) * background[1]) + (rgba[3] * rgba[1])),
        round(((1 - rgba[3]) * background[2]) + (rgba[3] * rgba[2])),
    )
m0h17
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1

Here is a convenient SASS function in accordance with Andras' and hkurabko's answers.

@function rgba_blend($fore, $back) {
  $ored: ((1 - alpha($fore)) * red($back) ) + (alpha($fore) * red($fore));
  $ogreen: ((1 - alpha($fore)) * green($back) ) + (alpha($fore) * green($fore));
  $oblue: ((1 - alpha($fore)) * blue($back) ) + (alpha($fore) * blue($fore));
  @return rgb($ored, $ogreen, $oblue);
}

Usage:

$my_color: rgba(red, 0.5); // build a color with alpha for below

#a_div {
  background-color: rgba_blend($my_color, white);
}
Rohanthewiz
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0

Typescript function in accordance with hkurabko's answer.

type RgbType = { red: number; green: number; blue: number; };
type RgbaType = { alpha: number; } & RgbType;

export const blendAlphaToBackground = (
  rgba: RgbaType,
  background: RgbaType = {red: 255, green: 255, blue: 255, alpha: 1},
): RgbaType => {
  const red = Math.round((1 - rgba.alpha) * background.red + rgba.alpha * rgba.red);
  const green = Math.round((1 - rgba.alpha) * background.green + rgba.alpha * rgba.green);
  const blue = Math.round((1 - rgba.alpha) * background.blue + rgba.alpha * rgba.blue);

  return {red, green, blue, alpha: 1};
};
A. Masson
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0

In my case I wanted to make the RGB image look as if it was an RGBA image on a white background. As typical converting methods just remove the A channel, this can result in pixels in the RGB channels to become visible that were previously made transparent by the alpha channel.

The following worked for me:

import numpy as np

def convert_RGBA_to_RGB(input_image): 
    # Takes an RGBA image as input
    
    # Based on the following chat with user Andras Deak
    ## https://chat.stackoverflow.com/transcript/message/55060299#55060299
   
    input_image_normed = input_image / 255  # shape (nx, ny, 4), dtype float
    alpha = input_image_normed[..., -1:]  # shape (nx, ny, 1) for broadcasting
    input_image_normed_rgb = input_image_normed[..., :-1]  # shape (nx, ny, 3)
    #bg_normed = np.zeros_like(red_normed_rgb)  # shape (nx, ny, 3) <-- black background
    bg_normed = np.ones_like(input_image_normed_rgb)  # shape (nx, ny, 3) <-- white background
    composite_normed = (1 - alpha) * bg_normed + alpha * input_image_normed_rgb
    composite = (composite_normed * 255).round().astype(np.uint8)
    return composite
Mitchell van Zuylen
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