Programmatically Lighten a Color

Question

Motivation

I'd like to find a way to take an arbitrary color and lighten it a few shades, so that I can programatically create a nice gradient from the one color to a lighter version. The gradient will be used as a background in a UI.

Possibility 1

Obviously I can just split out the RGB values and increase them individually by a certain amount. Is this actually what I want?

Possibility 2

My second thought was to convert the RGB to HSV/HSB/HSL (Hue, Saturation, Value/Brightness/Lightness), increase the brightness a bit, decrease the saturation a bit, and then convert it back to RGB. Will this have the desired effect in general?

If the background is white, then modifying alpha/transparency can often work well as a hack to "lighten" the color. — astreltsov, May 22 '19 at 13:14

score 61 · Answer 1 · edited May 23 '17 at 12:02

As Wedge said, you want to multiply to make things brighter, but that only works until one of the colors becomes saturated (i.e. hits 255 or greater). At that point, you can just clamp the values to 255, but you'll be subtly changing the hue as you get lighter. To keep the hue, you want to maintain the ratio of (middle-lowest)/(highest-lowest).

Here are two functions in Python. The first implements the naive approach which just clamps the RGB values to 255 if they go over. The second redistributes the excess values to keep the hue intact.

def clamp_rgb(r, g, b):
    return min(255, int(r)), min(255, int(g)), min(255, int(b))

def redistribute_rgb(r, g, b):
    threshold = 255.999
    m = max(r, g, b)
    if m <= threshold:
        return int(r), int(g), int(b)
    total = r + g + b
    if total >= 3 * threshold:
        return int(threshold), int(threshold), int(threshold)
    x = (3 * threshold - total) / (3 * m - total)
    gray = threshold - x * m
    return int(gray + x * r), int(gray + x * g), int(gray + x * b)

I created a gradient starting with the RGB value (224,128,0) and multiplying it by 1.0, 1.1, 1.2, etc. up to 2.0. The upper half is the result using clamp_rgb and the bottom half is the result with redistribute_rgb. I think it's easy to see that redistributing the overflows gives a much better result, without having to leave the RGB color space.

Lightness gradient with clamping (top) and redistribution (bottom)

For comparison, here's the same gradient in the HLS and HSV color spaces, as implemented by Python's colorsys module. Only the L component was modified, and clamping was performed on the resulting RGB values. The results are similar, but require color space conversions for every pixel.

Lightness gradient with HLS (top) and HSV (bottom)

After upvoting this I realized this is way overkill. All you need to do is alpha blend with white at a certain opacity (i.e. interpolate the color components the same fraction towards 255). — Andy, Jun 08 '15 at 22:50
@Andy take a look at the sample. The first two boxes on the left don't have any white in them, they're fully saturated. It would have been more if I had started with a darker color. — Mark Ransom, Jun 09 '15 at 02:30
Ah, I see, so your code has the same linear discontinuity as the HLS colorspace. I can see how that would make a richer gradient. Since I just wanted to lighten a color for a hover highlight instead of making a gradient, I was expecting a simpler method — Andy, Jun 09 '15 at 02:46
@Andy there's another answer that suggests a linear interpolation with white, I believe that's exactly what you're after. — Mark Ransom, Jun 09 '15 at 03:07
Yes, I was going to add that answer myself until I saw someone else had done it. Cheers! — Andy, Jun 09 '15 at 03:08

Grey Panther · Accepted Answer · 2013-01-21T13:15:07.410

39

I would go for the second option. Generally speaking the RGB space is not really good for doing color manipulation (creating transition from one color to an other, lightening / darkening a color, etc). Below are two sites I've found with a quick search to convert from/to RGB to/from HSL:

from the "Fundamentals of Computer Graphics"
some sourcecode in C# - should be easy to adapt to other programming languages.

edited Jan 21 '13 at 13:15

answered Sep 26 '08 at 20:44

Grey Panther

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Good replacement link for conversion source code: http://bobpowell.net/RGBHSB.aspx – Curtis Yallop Jul 26 '13 at 17:25
Another reference link: http://social.msdn.microsoft.com/Forums/vstudio/en-US/f05428dd-b471-4c47-b52e-55449b6bf6a1/rgbhsb-color-conversion – Curtis Yallop Jul 26 '13 at 17:28
Update 2021: The first link of Curtis Yallop no longer works and the second refers only to further links. So this is a step by step programming receipt: https://css-tricks.com/converting-color-spaces-in-javascript/ And here are wonderful insights what is behind the conversion of color spaces: https://stackoverflow.com/a/39147465/818827 - I ended up with the second option of the OP, after fiddling around with handy, simple approaches which have their unwelcome limits. – peter_the_oak Apr 03 '21 at 07:48

score 26 · Answer 3 · answered Sep 26 '08 at 20:46

26

In C#:

public static Color Lighten(Color inColor, double inAmount)
{
  return Color.FromArgb(
    inColor.A,
    (int) Math.Min(255, inColor.R + 255 * inAmount),
    (int) Math.Min(255, inColor.G + 255 * inAmount),
    (int) Math.Min(255, inColor.B + 255 * inAmount) );
}

I've used this all over the place.

answered Sep 26 '08 at 20:46

Nick

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Very handy, and simple. Using it in a VB.NET app now. – Kezzer Sep 16 '09 at 11:13
See http://stackoverflow.com/questions/801406/c-create-a-lighter-darker-color-based-on-a-system-color – Martin Marconcini Apr 22 '10 at 14:29
If you are using `using System.Windows.Media.Color;` shouldn't you be casting to byte ? – Devid Sep 10 '18 at 15:05

score 11 · Answer 4 · answered Oct 21 '08 at 10:45

ControlPaint class in System.Windows.Forms namespace has static methods Light and Dark:

public static Color Dark(Color baseColor, float percOfDarkDark);

These methods use private implementation of HLSColor. I wish this struct was public and in System.Drawing.

Alternatively, you can use GetHue, GetSaturation, GetBrightness on Color struct to get HSB components. Unfortunately, I didn't find the reverse conversion.

score 8 · Answer 5 · answered Sep 26 '08 at 20:47

8

Convert it to RGB and linearly interpolate between the original color and the target color (often white). So, if you want 16 shades between two colors, you do:


for(i = 0; i < 16; i++)
{
  colors[i].R = start.R + (i * (end.R - start.R)) / 15;
  colors[i].G = start.G + (i * (end.G - start.G)) / 15;
  colors[i].B = start.B + (i * (end.B - start.B)) / 15;
}

answered Sep 26 '08 at 20:47

Adam Rosenfield

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In most cases only the start and end color are needed since the graphics library of choice probably has built in gradient capabilities. – Max Schmeling Sep 26 '08 at 21:22
And by using integer division, you eliminate the need for round (floor). Clever. – Stefan Steiger Nov 19 '15 at 10:34

Pavel Vladov · Answer 6 · 2013-04-02T13:39:29.907

6

In order to get a lighter or a darker version of a given color you should modify its brightness. You can do this easily even without converting your color to HSL or HSB color. For example to make a color lighter you can use the following code:

float correctionFactor = 0.5f;
float red = (255 - color.R) * correctionFactor + color.R;
float green = (255 - color.G) * correctionFactor + color.G;
float blue = (255 - color.B) * correctionFactor + color.B;
Color lighterColor = Color.FromArgb(color.A, (int)red, (int)green, (int)blue);

If you need more details, read the full story on my blog.

edited Apr 02 '13 at 13:39

answered Oct 24 '12 at 08:34

Pavel Vladov

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+1, Nicks method, the correct way. But you can still skip the cast to and from float by doing it like Adam Rosenfield, by splitting the correction factor into steps and index, which also gives two understandable values instead of one abstract number between 0.0 and 1.0. e.g. your correction factor is Rosenfield's i/15, while assuming white as target color. – Stefan Steiger Nov 19 '15 at 10:45

score 4 · Answer 7 · answered Sep 26 '08 at 20:46

4

Converting to HS(LVB), increasing the brightness and then converting back to RGB is the only way to reliably lighten the colour without effecting the hue and saturation values (ie to only lighten the colour without changing it in any other way).

answered Sep 26 '08 at 20:46

David Arno

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score 3 · Answer 8 · edited May 23 '17 at 12:26

3

A very similar question, with useful answers, was asked previously: How do I determine darker or lighter color variant of a given color?

Short answer: multiply the RGB values by a constant if you just need "good enough", translate to HSV if you require accuracy.

edited May 23 '17 at 12:26

Community

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answered Sep 26 '08 at 20:45

Wedge

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Upvoted for the reference to the similar question, but I disagree with the RGB advice - go HSV. – Blair Conrad Sep 26 '08 at 20:55

user1873073 · Answer 9 · 2013-01-23T18:34:33.523

I used Andrew's answer and Mark's answer to make this (as of 1/2013 no range input for ff).

function calcLightness(l, r, g, b) {
    var tmp_r = r;
    var tmp_g = g;
    var tmp_b = b;

    tmp_r = (255 - r) * l + r;
    tmp_g = (255 - g) * l + g;
    tmp_b = (255 - b) * l + b;

    if (tmp_r > 255 || tmp_g > 255 || tmp_b > 255) 
        return { r: r, g: g, b: b };
    else 
        return { r:parseInt(tmp_r), g:parseInt(tmp_g), b:parseInt(tmp_b) }
}

enter image description here

score 2 · Answer 10 · answered Sep 26 '08 at 20:54

I've done this both ways -- you get much better results with Possibility 2.

Any simple algorithm you construct for Possibility 1 will probably work well only for a limited range of starting saturations.

You would want to look into Poss 1 if (1) you can restrict the colors and brightnesses used, and (2) you are performing the calculation a lot in a rendering.

Generating the background for a UI won't need very many shading calculations, so I suggest Poss 2.

-Al.

score 1 · Answer 11 · answered Dec 19 '13 at 16:53

Pretend that you alpha blended to white:

oneMinus = 1.0 - amount
r = amount + oneMinus * r
g = amount + oneMinus * g
b = amount + oneMinus * b

where amount is from 0 to 1, with 0 returning the original color and 1 returning white.

You might want to blend with whatever the background color is if you are lightening to display something disabled:

oneMinus = 1.0 - amount
r = amount * dest_r + oneMinus * r
g = amount * dest_g + oneMinus * g
b = amount * dest_b + oneMinus * b

where (dest_r, dest_g, dest_b) is the color being blended to and amount is from 0 to 1, with zero returning (r, g, b) and 1 returning (dest.r, dest.g, dest.b)

I think this should be the top answer. Likewise to darken you just alpha blend to black (i.e. multiply components by some `amount < 1`). — Andy, Jun 08 '15 at 22:51

score 1 · Answer 12 · edited May 23 '17 at 11:47

I didn't find this question until after it became a related question to my original question.

However, using insight from these great answers. I pieced together a nice two-liner function for this:

Programmatically Lighten or Darken a hex color (or rgb, and blend colors)

Its a version of method 1. But with over saturation taken into account. Like Keith said in his answer above; use Lerp to seemly solve the same problem Mark mentioned, but without redistribution. The results of shadeColor2 should be much closer to doing it the right way with HSL, but without the overhead.

score 1 · Answer 13 · answered Oct 21 '08 at 10:47

IF you want to produce a gradient fade-out, I would suggest the following optimization: Rather than doing RGB->HSB->RGB for each individual color you should only calculate the target color. Once you know the target RGB, you can simply calculate the intermediate values in RGB space without having to convert back and forth. Whether you calculate a linear transition of use some sort of curve is up to you.

score 1 · Answer 14 · edited May 23 '17 at 11:47

1

Method 1: Convert RGB to HSL, adjust HSL, convert back to RGB.

Method 2: Lerp the RGB colour values - http://en.wikipedia.org/wiki/Lerp_(computing)

See my answer to this similar question for a C# implementation of method 2.

edited May 23 '17 at 11:47

Community

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answered Apr 22 '10 at 11:57

Keith

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score 1 · Answer 15 · answered Apr 14 '18 at 19:19

1

A bit late to the party, but if you use javascript or nodejs, you can use tinycolor library, and manipulate the color the way you want:

tinycolor("red").lighten().desaturate().toHexString() // "#f53d3d"

answered Apr 14 '18 at 19:19

Ari

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score 0 · Answer 16 · answered Sep 26 '08 at 20:44

0

I would have tried number #1 first, but #2 sounds pretty good. Try doing it yourself and see if you're satisfied with the results, it sounds like it'll take you maybe 10 minutes to whip up a test.

answered Sep 26 '08 at 20:44

davr

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score 0 · Answer 17 · answered Sep 26 '08 at 20:56

Technically, I don't think either is correct, but I believe you want a variant of option #2. The problem being that taken RGB 990000 and "lightening" it would really just add onto the Red channel (Value, Brightness, Lightness) until you got to FF. After that (solid red), it would be taking down the saturation to go all the way to solid white.

The conversions get annoying, especially since you can't go direct to and from RGB and Lab, but I think you really want to separate the chrominance and luminence values, and just modify the luminence to really achieve what you want.

score 0 · Answer 18 · answered Jan 14 '10 at 14:11

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I have a blog post that shows how to do this in Delphi. Its pretty simple because the functions ColorRGBToHSL and ColorHLSToRGB are part of the standard library.

answered Jan 14 '10 at 14:11

Lawrence Barsanti

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score 0 · Answer 19 · answered Oct 21 '08 at 10:36

0

You'll find code to convert between color spaces in the color-tools ruby library

answered Oct 21 '08 at 10:36

Thibaut Barrère

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score 0 · Answer 20 · answered May 08 '17 at 01:49

Here's an example of lightening an RGB colour in Python:

def lighten(hex, amount):
    """ Lighten an RGB color by an amount (between 0 and 1),

    e.g. lighten('#4290e5', .5) = #C1FFFF
    """
    hex = hex.replace('#','')
    red = min(255, int(hex[0:2], 16) + 255 * amount)
    green = min(255, int(hex[2:4], 16) + 255 * amount)
    blue = min(255, int(hex[4:6], 16) + 255 * amount)
    return "#%X%X%X" % (int(red), int(green), int(blue))

I have absolutely no idea of what this does. – user2752471 Apr 21 '21 at 22:20 — user2752471, Apr 21 '21 at 22:20

Programmatically Lighten a Color

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