Algorithm: How do I fade from Red to Green via Yellow using RGB values?

Question

I want to display a color based on a value from 0 to 100. At one end (100), it's pure Red, the other end (0), pure Green. In the middle (50), I want it to be yellow.

And I want the colors to fade gradually from one to another, such that at 75, the color is half red and half yellow, etc.

How do I program the RGB values to reflect this fading? Thanks.

WPF? WinForms? What are you going to apply the gradient to? – Alex Aza Jun 18 '11 at 06:30 — Alex Aza, Jun 18 '11 at 06:30

score 100 · Answer 1 · edited Oct 15 '19 at 15:56

100

I had the same need and I just resolved with this:

myColor = new Color(2.0f * x, 2.0f * (1 - x), 0);

Explanation: Instead of the [0-255] range, let's focus on the [0.0-1.0] range for color components:

Green = 0.0, 1.0, 0.0
Yellow = 1.0, 1.0, 0.0
Red= 1.0, 0.0, 0.0

If you just scale the green component from 0.0 (on one end) to 1.0 (on the other end) and do the same thing with the red component (but going backwards), you'll get ugly and non-uniform color distribution.

To make it look nice, we could write a lot of code, or we could be more clever.

If you look carefully at the single components, you can see that we can split the range in two equal parts: in the first one we increase the red component from 0.0 to 1.0, leaving the green at 1.0 and the blue at 0.0; in the second we decrease the green component, leaving the other 2 as they are. We can take advantage of the fact that any value above 1.0 will be read as 1.0, by maxing out our values to simplify the code. Assuming your x value goes from 0.00 (0%) to 1.00 (100%), you can multiply it by 2 to let it go over the 1.0 limit for color components. Now you have your components going from 0.0 to 2.0 (the red one) and from 2.0 to 0.0 (the green one). Let them be clipped to [0.0-1.0] ranges and there you go.

If your x moves in another range (like [0-100]) you need to choose an appropriate factor instead of 2

edited Oct 15 '19 at 15:56

BenV

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answered Oct 30 '11 at 21:05

Giorgio Aresu

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Thank you very much. I thought of this approach when I found myself in this exact situation, because other solutions seemed too complicate for such a task – Giorgio Aresu Mar 02 '14 at 01:04
This is definitely the best answer as it is simple and works perfectly. I actually had to work out a few values to understand fully why it worked so well – Felipe Nov 25 '14 at 04:47
2

Excellent solution for our needs. We're computing it in Javascript and going from Red through White to Green. This adds an interesting twist as the Yellow component goes from 0.0 to 1.0 and back to 0.0 through the scale. This computes as yellowValue = 1 - Math.abs(value - 0.5) * 2; – LarzStarz Sep 18 '15 at 14:09
Great idea @GiorgioAresu! This is one of those answers that I believe deserves far more up votes. Great solution, and great write up to explain the idea. – Adam Hopkins Jun 01 '16 at 04:06
4

@GiorgioAresu, I took your principle and applied it to a utility on a codepen found here: http://codepen.io/ahopkins/full/beNWWp/ – Adam Hopkins Jun 01 '16 at 06:05
5

Works perfectly! I was using python (PyQt5) and had to modify slightly: `QColor(min(255,2*255*(temp)),min(255,2*255*(1-temp)),0)` – laserpython Aug 01 '17 at 21:38
In JS i adapted it to this: ```const red = Math.min(2 - (2 * Fraction), 1) * 255; const green = Math.min(2 * Fraction, 1) * 255;``` – Ansjovis86 Feb 02 '18 at 14:16
This solution doesn't go through yellow. It's a nice solution if you need green to red gradient. Strictly speaking, it doesn't answer OPs question. – duedl0r Nov 25 '19 at 10:14
@duedl0r yes it does, as explained in the answer. It does so while keeping the yellow at the exact center of the gradient, instead of skewed at one side. Did you try it? – Giorgio Aresu Nov 26 '19 at 13:17
@GiorgioAresu: Yes you're right. I tried it, but made an error: I didn't read to the end ;) – duedl0r Nov 27 '19 at 01:14
additionally, instead of using `if` statements to bound your ranges, use `%`. No branching in your code. – Hanif Bin Ariffin Feb 21 '20 at 03:43
@HanifBinAriffin I think you meant to reply to a different answer :) – Giorgio Aresu Feb 22 '20 at 06:40
@AdamHopkins I thnk your codepen has a bug in the clip function or at least I'm readingit as `if (condition) return X else return X // also` – netizen Dec 13 '22 at 17:51

score 50 · Accepted Answer · answered Jun 18 '11 at 06:35

50

The RGB values for the colors:

Red 255, 0, 0
Yellow 255, 255, 0
Green 0, 255, 0

Between Red and Yellow, equally space your additions to the green channel until it reaches 255. Between Yellow and Green, equally space your subtractions from the red channel.

answered Jun 18 '11 at 06:35

jterrace

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Colors are not linear, a change from 0 to 64 I'd bigger than a change from 191 to 255, this is especially clear on led displays, as their black is blacker. – Ferrybig Oct 15 '19 at 16:16
2

@Ferrybig, taking your comment into consideration, what is the proper interpolation formula? – joejoejoejoe4 Apr 04 '20 at 21:53

score 20 · Answer 3 · edited Oct 03 '17 at 12:36

20

Here is a very simple linear interpolation of the color components. It might serve your needs.

public Color GetBlendedColor(int percentage)
{
    if (percentage < 50)
        return Interpolate(Color.Red, Color.Yellow, percentage / 50.0);
    return Interpolate(Color.Yellow, Color.Lime, (percentage - 50) / 50.0);
}

private Color Interpolate(Color color1, Color color2, double fraction)
{
    double r = Interpolate(color1.R, color2.R, fraction);
    double g = Interpolate(color1.G, color2.G, fraction);
    double b = Interpolate(color1.B, color2.B, fraction);
    return Color.FromArgb((int)Math.Round(r), (int)Math.Round(g), (int)Math.Round(b));
}

private double Interpolate(double d1, double d2, double fraction)
{
    return d1 + (d2 - d1) * fraction;
}

edited Oct 03 '17 at 12:36

Fidel

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answered Jun 18 '11 at 06:45

Rick Sladkey

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In the Interpolate function, assuming d2 is your target color and d1 is your begin color, it should be (d2 - d1) instead of (d1 - d2). – Tincan Jan 06 '16 at 11:23
This helped me a lot with custom colors and the fix to `d1` and `d2`. – Christian Ivicevic Jan 08 '17 at 21:13

score 17 · Answer 4 · edited Oct 03 '17 at 12:48

I don't know C#, so this answer is just a suggested approach. Let x denote the int that ranges from 0 to 100. Something like this should work:

red   = (x > 50 ? 1-2*(x-50)/100.0 : 1.0);
green = (x > 50 ? 1.0 : 2*x/100.0);
blue  = 0.0

The idea is to start at red: (1.0,0.0,0.0). Then increase the green to get yellow: (1.0,1.0,0.0). Then decrease the red to get green: (0.0,1.0,0.0).

Edit: Here is the code in C#

static Color GetColorFromRedYellowGreenGradient(double percentage)
{
    var red = (percentage > 50 ? 1 - 2 * (percentage - 50) / 100.0 : 1.0) * 255;
    var green = (percentage > 50 ? 1.0 : 2 * percentage / 100.0) * 255;
    var blue = 0.0;
    Color result = Color.FromArgb((int)red, (int)green, (int)blue);
    return result;
}

Voxpire · Answer 5 · 2016-10-18T22:20:47.367

Simplified extension method;

public static Color Interpolate(this Color source, Color target, double percent)
{
    var r = (byte)(source.R + (target.R - source.R) * percent);
    var g = (byte)(source.G + (target.G - source.G) * percent);
    var b = (byte)(source.B + (target.B - source.B) * percent);

    return Color.FromArgb(255, r, g, b);
}

Usage;

var low = 33.0;
var high = 100.0;
var color = Color.Red.Interpolate(Color.Green, low / high);

score 6 · Answer 6 · answered Sep 16 '16 at 22:42

6

After I experimented for a while with more realistic colors, here is my formula:

public Color GetColorOf(double value, double minValue, double maxValue)
{
    if (value == 0 || minValue == maxValue) return Color.White;

    var g = (int)(240 * value / maxValue);
    var r = (int)(240 * value / minValue);

    return (value > 0
        ? Color.FromArgb(240 - g, 255 - (int)(g * ((255 - 155) / 240.0)), 240 - g)
        : Color.FromArgb(255 - (int)(r * ((255 - 230) / 240.0)), 240 - r, 240 - r));
}

You get no background (i.e. Color.White) for 0 (or NULL), or when min = max.
For all positive values, you get an evenly distributed green color between RGB(240, 255, 240) and RGB(0, 155, 0).
For all negative values, you get an evenly distributed red color between RGB(255, 240, 240) and RGB(230, 0, 0).

answered Sep 16 '16 at 22:42

Cristian Scutaru

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This doesn't answer the original question. Not only does it go through white instead of yellow, it operates on a different range of numbers than requested. – Setsu Sep 16 '16 at 23:02
1

+1 for having a green-to-white-to-red algorithm which fits me, but indeed does not answer the OP question. – Benj Nov 15 '16 at 14:42
1

Don't care what question this answers... this is genius. – Hambone Apr 04 '18 at 03:44

score 5 · Answer 7 · edited May 23 '17 at 10:31

5

You need to use the HSB or HSV color representation instead, and play with the H ("Hue") value. See this other SO question for transformation betweeen RGB and HSB/HSV: How to change RGB color to HSV?

edited May 23 '17 at 10:31

Community

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answered Jun 18 '11 at 06:37

Simon Mourier

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score 1 · Answer 8 · answered Feb 15 '18 at 17:16

I wrote a Swift 4 extension for UIColor which converts a percentage (0-100) to a UIColor from red to green.

I'm currently using it for progress bars in an analytics app. Here's an example:

let myNewColor = UIColor().greenRedProgress(percent: Int))

Here's the extension:

extension UIColor{

func greenRedProgress(percent: Int) -> UIColor{
    let modVal = CGFloat((Double(percent).truncatingRemainder(dividingBy: 50) / 50) * 255)
    if percent <= 0{
        return UIColor(red: 1.0, green: 0, blue: 0, alpha: 1)
    }else if percent >= 100{
        return UIColor(red: 0, green: 1.0, blue: 0, alpha: 1)
    }else{
        switch percent{
            case 1..<50: return UIColor(red: 1.0, green: (modVal/255), blue: 0, alpha: 1)
            case 51..<100: return UIColor(red: (255 - modVal)/255, green: 1.0, blue: 0, alpha: 1)
            case 50: return UIColor(red: 1.0, green: 1.0, blue: 0, alpha: 1)
            default: return UIColor(red: 0, green: 1.0, blue: 0, alpha: 1)
        }
    }
}}

score 1 · Answer 9 · answered Dec 19 '18 at 19:28

This method (in c#, but can be easily translated to other languages) will take a percentage and list of colors and return the color on the gradient based on your percentage. When you pass in the colors, they need to be in order from 0 value to 100 value (so you would want to pass Green, Yellow, Red - in that order). If you find that you want different or more colors in the middle, just add them to the list of colors you pass in the order you want them to appear.

public Color ColorBasedOnPercent(decimal percent, params Color[] colors)
    {
        if (colors.Length == 0)
        {
            //I am using Transparent as my default color if nothing was passed
            return Color.Transparent;
        }
        if (percent > 1)
        {
            percent = percent / 100;
        }

        //find the two colors within your list of colors that the percent should fall between
        var colorRangeIndex = (colors.Length - 1) * percent;
        var minColorIndex = (int)Math.Truncate(colorRangeIndex);
        var maxColorIndex = minColorIndex + 1;
        var minColor = colors[minColorIndex];

        if (maxColorIndex < colors.Length)
        {
            var maxColor = colors[maxColorIndex];

            //get the differences between all the color values for the two colors you are fading between
            var aScale = maxColor.A - minColor.A;
            var redScale = maxColor.R - minColor.R;
            var greenScale = maxColor.G - minColor.G;
            var blueScale = maxColor.B - minColor.B;

            //the decimal distance of how "far" this color should be from the minColor in the range
            var gradientPct = colorRangeIndex - minColorIndex;

            //for each piece of the color (ARGB), add a percentage(gradientPct) of the distance between the two colors
            int getRGB(int originalRGB, int scale) => (int)Math.Round(originalRGB + (scale * gradientPct));

            return Color.FromArgb(getRGB(minColor.A, aScale), getRGB(minColor.R, redScale), getRGB(minColor.G, greenScale), getRGB(minColor.B, blueScale));
        }
        return minColor;
    }

score 1 · Answer 10 · answered Jun 18 '11 at 06:34

1

Take a look at LinearGradientBrush. It should be a complete implementation on what you're looking for.

answered Jun 18 '11 at 06:34

PedroC88

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score 0 · Answer 11 · answered Jul 16 '13 at 18:36

0

you just need to create a function with integer parameter

input 100 will return RGB (100, 0, 0)
input 50 will return RGB (50, 50, 0)
input 0 will return RGB (0, 100, 0)
input 99 will return RGB (99, 1, 0)
input 98 will return RGB (98, 2, 0)
input 2 will return RGB (2, 98, 0)

input 1 will return RGB (1, 99, 0)

    private Color fader(int v){
       return Color.FromArgb(v, 100-v, 0); 
    }

answered Jul 16 '13 at 18:36

Tun Zarni Kyaw

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It looks like you forgot that `r.g.b` values are up to 255 and not 100. – Rafael Herscovici Feb 17 '15 at 09:21
This won't actually fade nicely, the yellow section will be pretty narrow – Giorgio Aresu Dec 29 '16 at 09:53

score 0 · Answer 12 · answered Oct 17 '14 at 22:11

I had a need for something similar today. Input was percent from 0.0 to 1.0, and output red to green. Implementation based on jterrace's answer:

Color percentToColor(float percent)
{
    if (percent<0 || percent>1) { return Color.Black; }

    int r, g;
    if (percent<0.5)
    {
        r=255;
        g = (int)(255*percent/0.5);  //closer to 0.5, closer to yellow (255,255,0)
    }
    else
    {
        g=255;
        r = 255 - (int)(255*(percent-0.5)/0.5); //closer to 1.0, closer to green (0,255,0)
    }
    return Color.FromArgb(r, g, 0);
}

score 0 · Answer 13 · answered Oct 22 '17 at 11:43

I mapped values of CPU speed between 1400 MHz and 3500 MHz to rgb() values to get from green -> yellow -> red with this function

function green_yellow_red(core_MHz, core_id){
  var core_color = ~~core_MHz.map(1400, 3500, 0, 510)

  if(core_color < 255){
    $('#cpu_core_'+core_id).css('background', 'rgb('+core_color+',255 , 0)')
  }else{
    core_color-=255
    $('#cpu_core_'+core_id).css('background', 'rgb(255 ,'+ (255-core_color) +', 0)')
  }
}

score 0 · Answer 14 · answered Mar 19 '19 at 15:31

More of the same. Just Delphi Pascal coded and simplified+locked to semaphore colors (red/yellow/green).

rectangle1.Fill.Color:=DefineColorSemaphore(newcolor);

function TForm1.DefineColorSemaphore(valperc:integer):TAlphaColor;
var
   vcol: TAlphaColorRec;
begin
  vcol.B := 0;    // blue:  always 0
  vcol.A := 255;  // alpha: 255=no
  if (valperc < 50) then
    begin
       // starts @ RGB=255,0,0 and increases G 0->255
       vcol.R := 255;
       vcol.G := trunc(255*valperc/50);
    end
    else
    begin
       // starts @ RGB=255,255,0 and decreases R 255->0
       vcol.R := 255-trunc(255* (valperc - 50)/50);
       vcol.G := 255;
    end;
  result:= TAlphaColor(vcol);
end;

avijit · Answer 15 · 2019-11-16T08:24:33.120

Recently I was planning to do samething using Javascript. I have followed two steps.

First scale the value in 0-1 range(scaledValue = (value - min) / (max - min))
Then increase value of Red channel if the value is less than or equal 0.5 and then decrease value of Green channel if the value is greater than 0.5

Here is the code I used for this purpose.

var blue = 0.0, red = 0.0, green = 0.0;

if(scaledValue <= 0.5)
{
    red = (scaledValue * 2) * 255.0;
    green = 255.0;
    blue = 0;
}else
{
    red = 255.0;
    green = 255.0 + 255.0 - ((scaledValue  * 2)* 255);
    blue = 0;
}

A sample code for showing this works in C++

This solution is inspired by @jterrace answer.

score 0 · Answer 16 · answered Jun 26 '20 at 15:24

// To fill an OwnerDraw control with red-yellow-green gradient upon WM_DRAWITEM

int cx = lpDis->rcItem.right - lpDis->rcItem.left;

for (int x = 0; x < cx; x++)
{
    COLORREF cr;

    double d = 255 * 2 * (double)x/(double)cx;

    cr = x <= cx/2 ?    RGB(255,     d,   0) :
                        RGB(255 - d, 255, 0);

    HPEN hPen = CreatePen(PS_SOLID, 1, cr);
    HPEN hOldPen = (HPEN)SelectObject(lpDis->hDC, hPen);
    MoveToEx(lpDis->hDC, x, lpDis->rcItem.top, NULL);
    LineTo(lpDis->hDC, x, lpDis->rcItem.bottom);
    SelectObject(lpDis->hDC, hOldPen);
    DeleteObject(hPen);
}

Illustration

AssaultOPS · Answer 17 · 2021-05-20T01:57:27.363

If you are on python then this might help...


def rgb(p):# <-- percentage as parameter
    #Starting with color red
    d = [255,0,0]
    #formula for finding green value by percentage
    d[1] = int((510*p)/100)
    print(d[1])
    #if green value more than 255
    #set green value 255
    #reduce the red value from remaining green value
    if d[1]>255:
        d[0] -= d[1]-255
        d[1] = 255
        
    return d

print(rgb(0))

Algorithm: How do I fade from Red to Green via Yellow using RGB values?

17 Answers17

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