Generate colors between red and green for a power meter?

Language AgnosticGraphicsColorsInterpolation

Language Agnostic Problem Overview


I'm writing a Java game and I want to implement a power meter for how hard you are going to shoot something.

I need to write a function that takes a int between 0 - 100, and based on how high that number is, it will return a color between Green (0 on the power scale) and Red (100 on the power scale).

Similar to how volume controls work:
volume control

What operation do I need to do on the Red, Green, and Blue components of a color to generate the colors between Green and Red?

So, I could run say, getColor(80) and it will return an orangish color (its values in R, G, B) or getColor(10) which will return a more Green/Yellow RGB value.

I know I need to increase components of the R, G, B values for a new color, but I don't know specifically what goes up or down as the colors shift from Green-Red.


Progress:

I ended up using HSV/HSB color space because I liked the gradiant better (no dark browns in the middle).

The function I used was:

public Color getColor(double power)
{
    double H = power * 0.4; // Hue (note 0.4 = Green, see huge chart below)
    double S = 0.9; // Saturation
    double B = 0.9; // Brightness

    return Color.getHSBColor((float)H, (float)S, (float)B);
}

Where "power" is a number between 0.0 and 1.0. 0.0 will return a bright red, 1.0 will return a bright green.

Java Hue Chart:
Java Hue Chart

Language Agnostic Solutions


Solution 1 - Language Agnostic

This should work - just linearly scale the red and green values. Assuming your max red/green/blue value is 255, and n is in range 0 .. 100

R = (255 * n) / 100
G = (255 * (100 - n)) / 100 
B = 0

(Amended for integer maths, tip of the hat to Ferrucio)

Another way to do would be to use a HSV colour model, and cycle the hue from 0 degrees (red) to 120 degrees (green) with whatever saturation and value suited you. This should give a more pleasing gradient.

Here's a demonstration of each technique - top gradient uses RGB, bottom uses HSV:

http://i38.tinypic.com/29o0q4k.jpg

Solution 2 - Language Agnostic

Off the top of my head, here is the green-red hue transition in HSV space, translated to RGB:

blue = 0.0
if 0<=power<0.5:        #first, green stays at 100%, red raises to 100%
    green = 1.0
    red = 2 * power
if 0.5<=power<=1:       #then red stays at 100%, green decays
    red = 1.0
    green = 1.0 - 2 * (power-0.5)

The red, green, blue values in the above example are percentages, you'd probably want to multiply them by 255 to get the most used 0-255 range.

Solution 3 - Language Agnostic

Short Copy'n'Paste answer...

On Java Std:

int getTrafficlightColor(double value){
    return java.awt.Color.HSBtoRGB((float)value/3f, 1f, 1f);
}

On Android:

int getTrafficlightColor(double value){
    return android.graphics.Color.HSVToColor(new float[]{(float)value*120f,1f,1f});
}

note: value is a number between 0 and 1 indicating the red-to-green condition.

Solution 4 - Language Agnostic

If you want an green-yellow-red representation like the accepted answer suggests then take a look at this.

http://jsfiddle.net/0awncw5u/2/

function percentToRGB(percent) {
    if (percent === 100) {
        percent = 99
    }
    var r, g, b;

    if (percent < 50) {
        // green to yellow
        r = Math.floor(255 * (percent / 50));
        g = 255;

    } else {
        // yellow to red
        r = 255;
        g = Math.floor(255 * ((50 - percent % 50) / 50));
    }
    b = 0;

    return "rgb(" + r + "," + g + "," + b + ")";
}


function render(i) {
    var item = "<li style='background-color:" + percentToRGB(i) + "'>" + i + "</li>";
    $("ul").append(item);
}

function repeat(fn, times) {
    for (var i = 0; i < times; i++) fn(i);
}


repeat(render, 100);

li {
    font-size:8px;
    height:10px;
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.0/jquery.min.js"></script>
<ul></ul>

Solution 5 - Language Agnostic

Linearly interpolating between green and red almost should work, except that in the middle there will be muddy brown color.

The most flexible and best looking solution is to have an image file somewhere that has the exact color ramp you want. And then lookup the pixel value in there. This has the flexibility that you can tweak the gradient to be just right.

If you still want to do it from code, then it's probably best to interpolate between green and yellow colors in the left side, and between yellow and red on the right side. In RGB space, green is (R=0, G=255, B=0), yellow is (R=255, G=255, B=0), red is (R=255, G=0, B=0) - this is assuming each color component goes from 0 to 255.

Solution 6 - Language Agnostic

I made a small function wich gives you the rgb integer value for a percentage value:

private int getGreenToRedGradientByValue(int currentValue, int maxValue)
{
	int r = ( (255 * currentValue) / maxValue );
	int g = ( 255 * (maxValue-currentValue) ) / maxValue;
	int b = 0;
	return ((r&0x0ff)<<16)|((g&0x0ff)<<8)|(b&0x0ff);
}

So if your currentValue is 50 and your maxValue is 100, this function will return the color that you need, so if you loop this function with a percentage value, your color value will go from green to red. Sorry for the bad explanation

Solution 7 - Language Agnostic

I'd say you want something in between a line segment in the HSV space (which has a slightly-too-bright yellow in the centre) and a line segment in the RGB space (which has an ugly brown in the centre). I would use this, where power = 0 will give green, power = 50 will give a slightly dull yellow, and power = 100 will give red.

blue = 0;
green = 255 * sqrt( cos ( power * PI / 200 ));
red = 255 * sqrt( sin ( power * PI / 200 )); 

Solution 8 - Language Agnostic

The accepted answer didn't even really answer the question...

C++

Here's a simple C++ code segment from my engine that linearly and efficiently interpolates between three arbitrary colors:

const MATH::FLOAT4 color1(0.0f, 1.0f, 0.0f, 1.0f);	// Green
const MATH::FLOAT4 color2(1.0f, 1.0f, 0.0f, 1.0f);	// Yellow
const MATH::FLOAT4 color3(1.0f, 0.0f, 0.0f, 1.0f);	// Red

MATH::FLOAT4 get_interpolated_color(float interpolation_factor)
{
	const float factor_color1 = std::max(interpolation_factor - 0.5f, 0.0f);
	const float factor_color2 = 0.5f - fabs(0.5f - interpolation_factor);
	const float factor_color3 = std::max(0.5f - interpolation_factor, 0.0f);

	MATH::FLOAT4 color;

	color.x = (color1.x * factor_color1 +
			   color2.x * factor_color2 +
			   color3.x * factor_color3) * 2.0f;

	color.y = (color1.y * factor_color1 +
			   color2.y * factor_color2 +
			   color3.y * factor_color3) * 2.0f;

	color.z = (color1.z * factor_color1 +
			   color2.z * factor_color2 +
			   color3.z * factor_color3) * 2.0f;

	color.w = 1.0f;

	return(color);
}

The interpolation_factor is assumed to be in the range of 0.0 ... 1.0.
The colors are assumed to be in the range of 0.0 ... 1.0 (e.g. for OpenGL).

C#

Here's the same function written in C#:

private readonly Color mColor1 = Color.FromArgb(255, 0, 255, 0);
private readonly Color mColor2 = Color.FromArgb(255, 255, 255, 0);
private readonly Color mColor3 = Color.FromArgb(255, 255, 0, 0);

private Color GetInterpolatedColor(double interpolationFactor)
{
    double interpolationFactor1 = Math.Max(interpolationFactor - 0.5, 0.0);
    double interpolationFactor2 = 0.5 - Math.Abs(0.5 - interpolationFactor);
    double interpolationFactor3 = Math.Max(0.5 - interpolationFactor, 0.0);

	return (Color.FromArgb(255,
                (byte)((mColor1.R * interpolationFactor1 +
                        mColor2.R * interpolationFactor2 +
                        mColor3.R * interpolationFactor3) * 2.0),

                (byte)((mColor1.G * interpolationFactor1 +
                        mColor2.G * interpolationFactor2 +
                        mColor3.G * interpolationFactor3) * 2.0),

                (byte)((mColor1.B * interpolationFactor1 +
                        mColor2.B * interpolationFactor2 +
                        mColor3.B * interpolationFactor3) * 2.0)));
}

The interpolationFactor is assumed to be in the range of 0.0 ... 1.0.
The colors are assumed to be in the range of 0 ... 255.

Solution 9 - Language Agnostic

You need to linearly interpolate (LERP) the color components. Here's how it's done in general, given a start value v0, an end value v1 and the required ratio (a normalized float between 0.0 and 1.0):

v = v0 + ratio * (v1 - v0)

This gives v0 when ratio is 0.0, v1 when ratio is 1.0, and everything between in the other cases.

You can do this either on the RGB components, or using some other color scheme, like HSV or HLS. The latter two will be more visually pleasing, since they work on hue and brightness compoments that map better to our color perception.

Solution 10 - Language Agnostic

Here is copy-and-paste solution for Swift and HSV scale:

UIColor initializer accepts hue, saturation and brightness in [0, 1] so for given value from [0, 1] we have:

let hue:        CGFloat = value / 3
let saturation: CGFloat = 1 // Or choose any
let brightness: CGFloat = 1 // Or choose any
let alpha:      CGFloat = 1 // Or choose any

let color = UIColor(hue: hue, saturation: saturation, brightness: brightness, alpha: alpha)

Solution 11 - Language Agnostic

import java.awt.Color;

public class ColorUtils {
	
	public static Color interpolate(Color start, Color end, float p) {
		float[] startHSB = Color.RGBtoHSB(start.getRed(), start.getGreen(), start.getBlue(), null);
		float[] endHSB = Color.RGBtoHSB(end.getRed(), end.getGreen(), end.getBlue(), null);
		
		float brightness = (startHSB[2] + endHSB[2]) / 2;
		float saturation = (startHSB[1] + endHSB[1]) / 2;
		
		float hueMax = 0;
		float hueMin = 0;
		if (startHSB[0] > endHSB[0]) {
			hueMax = startHSB[0];
			hueMin = endHSB[0];
		} else {
			hueMin = startHSB[0];
			hueMax = endHSB[0];
		}
		
		float hue = ((hueMax - hueMin) * p) + hueMin;
		
		return Color.getHSBColor(hue, saturation, brightness);
	}
}

Solution 12 - Language Agnostic

In Python 2.7:

import colorsys

def get_rgb_from_hue_spectrum(percent, start_hue, end_hue):
    # spectrum is red (0.0), orange, yellow, green, blue, indigo, violet (0.9)
    hue = percent * (end_hue - start_hue) + start_hue
    lightness = 0.5
    saturation = 1
    r, g, b = colorsys.hls_to_rgb(hue, lightness, saturation)
    return r * 255, g * 255, b * 255

# from green to red:
get_rgb_from_hue_spectrum(percent, 0.3, 0.0)

# or red to green:
get_rgb_from_hue_spectrum(percent, 0.0, 0.3)

Percent is of course value / max_value. Or if your scale doesn't begin at 0 then (value - min_value) / (max_value - min_value).

Solution 13 - Language Agnostic

If you need this kind of gradient (actually reversed) in CSS (min version 2.1) you can use HSL, too.

Supposing you are in an AngularJs template and value is a number from 0 to 1 and you want to style an element (background or text color)...

hsl({{ value * 120}}, 50%, 35%)

First value: degrees ( 0 red, 120 green) Second value: saturation (50% is neutral) Third value: lightness (50% neutral)

A nice article here

Theory on Wikipedia here

Solution 14 - Language Agnostic

Here is the Objective-C version of Simucal's solution:

- (UIColor*) colorForCurrentLevel:(float)level
{
    double hue = level * 0.4; // Hue (note 0.4 = Green)
    double saturation = 0.9; // Saturation
    double brightness = 0.9; // Brightness
    
    return [UIColor colorWithHue:hue saturation:saturation brightness:brightness alpha:1.0];
}

Where level would be a value between 0.0 and 1.0.
Hope this helps someone!

Solution 15 - Language Agnostic

JavaScript

I am using Google Visualization with bar chart and wanted the bars to be green to red based on a percentage. This turned out to be the cleanest solution I found and works perfectly.

function getGreenToRed(percent){
    r = percent<50 ? 255 : Math.floor(255-(percent*2-100)*255/100);
    g = percent>50 ? 255 : Math.floor((percent*2)*255/100);
    return 'rgb('+r+','+g+',0)';
}

Just make sure your percentage is 50 for 50% and not 0.50.

Solution 16 - Language Agnostic

I have updated the accepted answer by spliting it into first and second half of the (0,100) range and substituting 100 by a max level, so that the range can become dynamic. The result is exactly as with the HSV model, but still using RGB. The key is to have the (255,255,0) in the middle to represent the yellow colour. I have combined this idea in the accepted answer and another VBA code, so achieve it in VBA and use in Excel. I hope the logic helps and can be used in other languages/ applications.

Sub UpdateConditionalFormatting(rng As Range)
    Dim cell As Range
    Dim max As Integer

    max = WorksheetFunction.max(rng)

    For Each cell In rng.Cells
   
    If cell.Value >= 0 And cell.Value < max / 2 Then
        cell.Interior.Color = RGB(255 * cell.Value / (max / 2), 255, 0)
    ElseIf cell.Value >= max / 2 And cell.Value <= max Then
        cell.Interior.Color = RGB(255, 255 * ((max) - cell.Value) / (max / 2), 0)
    End If

    Next cell
End Sub

Cheers, Pavlin

Solution 17 - Language Agnostic

VB

Public Function GetPercentageColor( _
  ByVal iPercent As Long, Optional _
  ByVal bOpposit As Boolean) As Long
' 0->100% - Green->Yellow->Red
' bOpposit - Red->Yellow->Green

If bOpposit Then iPercent = (100 - iPercent)

Select Case iPercent
Case Is < 1: GetPercentageColor = 65280 ' RGB(0, 255, 0)
Case Is > 99: GetPercentageColor = 255  ' RGB(255, 0, 0)
Case Is < 50: GetPercentageColor = RGB(255 * iPercent / 50, 255, 0)
Case Else: GetPercentageColor = RGB(255, (255 * (100 - iPercent)) / 50, 0)
End Select

End Function

Solution 18 - Language Agnostic

Self contained example

<html>
<head>
<script>
//--------------------------------------------------------------------------
function gradient(left, mid, right)
{
    var obj = {}

    var lt50 = {"r":(mid.r-left.r)/50.0,
                "g":(mid.g-left.g)/50.0,
                "b":(mid.b-left.b)/50.0}
    var gt50 = {"r":(right.r-mid.r)/50.0,
                "g":(right.g-mid.g)/50.0,
                "b":(right.b-mid.b)/50.0}

    obj.getColor = function(percent) {
        if (percent == 50.0) {
            return mid;
        }
        if (percent < 50.0) {
            return "rgb("+Math.floor(left.r+lt50.r*percent+0.5)+","+
                          Math.floor(left.g+lt50.g*percent+0.5)+","+
                          Math.floor(left.b+lt50.b*percent+0.5)+")";
        }
        var p2 = percent-50.0;
        return "rgb("+Math.floor(mid.r+gt50.r*p2+0.5)+","+
                      Math.floor(mid.g+gt50.g*p2+0.5)+","+
                      Math.floor(mid.b+gt50.b*p2+0.5)+")";
    }

    return obj;
}

//--------------------------------------------------------------------------
var g_gradient = gradient( {"r":255, "g":20, "b":20},  // Left is red
                           {"r":255, "g":255, "b":20}, // Middle is yellow
                           {"r":20, "g":255, "b":20} ); // right is green

//--------------------------------------------------------------------------
function updateColor()
{
    var percent = document.getElementById('idtext').value.length;
    var oscore = document.getElementById('idscore');

    if (percent > 100.0) {
        percent = 100.0;
    }
    if (percent < 0.0) {
        percent = 0.0;
    }
    var col = g_gradient.getColor(percent)
    oscore.style['background-color'] = col;
    oscore.innerHTML = percent + '%';
}

</script>
</head>
<body onLoad="updateColor()">
<input size='100' placeholder='type text here' id='idtext' type="text" oninput="updateColor()" />
<br />
<br />
<div id='idscore' style='text-align:center; width:200px; border-style:solid;
     border-color:black; border-width:1px; height:20px;'> </div>
</body>
</html>

Solution 19 - Language Agnostic

This will vary from Red to Yellow and then to Green
value varies from 0 - 100

-(UIColor*) redToGreenColorWithPosition:(int) value {

    double R, G;
    if (value > 50) {
        R = (255 * (100 - value)/ 50) ;
        G = 255;
    }else {
        R = 255;
        G = (255 * (value*2)) / 100;
    }

    return [UIColor colorWithRed:R/255.0f green:G/255.0f blue:0.0f alpha:1.0f];
}


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QuestionmmcdoleView Question on Stackoverflow
Solution 1 - Language AgnosticPaul DixonView Answer on Stackoverflow
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