CODEWITHSUNDEEP
c#.netcolorssystem.drawing.color
Graviton
Graviton

Reputation: 83254

Generate color gradient in C#

My question here is similar to the question here, except that I am working with C#.

I have two colors, and I have a predefine steps. How to retrieve a list of Colors that are the gradients between the two?

This is an approach that I tried, which didn't work:

int argbMax = Color.Chocolate.ToArgb();
int argbMin = Color.Blue.ToArgb();
var colorList = new List<Color>();

for(int i=0; i<size; i++)
{
    var colorAverage= argbMin + (int)((argbMax - argbMin) *i/size);
    colorList.Add(Color.FromArgb(colorAverage));
}

If you try the above code, you will find that a gradual increase in argb doesn't correspond to a visual gradual increase in the color.

Any idea on this?

Upvotes: 30

Views: 50722

Answers (7)

sharp12345
sharp12345

Reputation: 4498

I have tried multiple answers in this page, but their output has issues, some has result that is missing one color gradient at the end, some has issues with the first input color being included in the output result.

Finally, this is what worked for me in the end:

    public static List<Color> GetGradientColors(Color start, Color end, int steps)
    {
        return GetGradientColors(start, end, steps, 0, steps - 1);
    }

    public static List<Color> GetGradientColors(Color start, Color end, int steps, int firstStep, int lastStep)
    {
        var colorList = new List<Color>();
        if (steps <= 0 || firstStep < 0 || lastStep > steps - 1)
            return colorList;

        double aStep = (end.A - start.A) / (double)steps;
        double rStep = (end.R - start.R) / (double)steps;
        double gStep = (end.G - start.G) / (double)steps;
        double bStep = (end.B - start.B) / (double)steps;

        for (int i = firstStep+1; i <= lastStep; i++)
        {
            var a = start.A + (aStep * i);
            var r = start.R + (rStep * i);
            var g = start.G + (gStep * i);
            var b = start.B + (bStep * i);
            colorList.Add(Color.FromArgb((int)a, (int)r, (int)g, (int)b));
        }

        return colorList;
    }

Upvotes: 0

William
William

Reputation: 2191

Combining this answer with the idea from several other answers to use floating-point steps, here's a complete method snippet for stepping with floating point. (With integer stepping, I had been getting asymmetrical gradient colors in a 16-color gradient from blue to red.)

Important difference in this version: you pass the total number of colors you want in the returned gradient sequence, not the number of steps to take within the method implementation.

public static IEnumerable<Color> GetColorGradient(Color from, Color to, int totalNumberOfColors)
{
    if (totalNumberOfColors < 2)
    {
        throw new ArgumentException("Gradient cannot have less than two colors.", nameof(totalNumberOfColors));
    }

    double diffA = to.A - from.A;
    double diffR = to.R - from.R;
    double diffG = to.G - from.G;
    double diffB = to.B - from.B;

    var steps = totalNumberOfColors - 1;

    var stepA = diffA / steps;
    var stepR = diffR / steps;
    var stepG = diffG / steps;
    var stepB = diffB / steps;

    yield return from;

    for (var i = 1; i < steps; ++i)
    {
        yield return Color.FromArgb(
            c(from.A, stepA),
            c(from.R, stepR),
            c(from.G, stepG),
            c(from.B, stepB));

        int c(int fromC, double stepC)
        {
            return (int)Math.Round(fromC + stepC * i);
        }
    }

    yield return to;
}

Upvotes: 2

David M
David M

Reputation: 72880

You will have to extract the R, G, B components and perform the same linear interpolation on each of them individually, then recombine.

int rMax = Color.Chocolate.R;
int rMin = Color.Blue.R;
// ... and for B, G
var colorList = new List<Color>();
for(int i=0; i<size; i++)
{
    var rAverage = rMin + (int)((rMax - rMin) * i / size);
    var gAverage = gMin + (int)((gMax - gMin) * i / size);
    var bAverage = bMin + (int)((bMax - bMin) * i / size);
    colorList.Add(Color.FromArgb(rAverage, gAverage, bAverage));
}

Upvotes: 36

MartinCermak
MartinCermak

Reputation: 111

Use double instead of int:

double stepA = ((end.A - start.A) / (double)(steps - 1));
double stepR = ((end.R - start.R) / (double)(steps - 1));
double stepG = ((end.G - start.G) / (double)(steps - 1));
double stepB = ((end.B - start.B) / (double)(steps - 1));

and:

yield return Color.FromArgb((int)start.A + (int)(stepA * step),
                                            (int)start.R + (int)(stepR * step),
                                            (int)start.G + (int)(stepG * step),
                                            (int)start.B + (int)(stepB * step));

Upvotes: 5

jocull
jocull

Reputation: 21115

Oliver's answer was very close... but in my case some of my stepper numbers needed to be negative. When converting the stepper values into a Color struct my values were going from negative to the higher values e.g. -1 becomes something like 254. I setup my step values individually to fix this.

public static IEnumerable<Color> GetGradients(Color start, Color end, int steps)
{
    int stepA = ((end.A - start.A) / (steps - 1));
    int stepR = ((end.R - start.R) / (steps - 1));
    int stepG = ((end.G - start.G) / (steps - 1));
    int stepB = ((end.B - start.B) / (steps - 1));

    for (int i = 0; i < steps; i++)
    {
        yield return Color.FromArgb(start.A + (stepA * i),
                                    start.R + (stepR * i),
                                    start.G + (stepG * i),
                                    start.B + (stepB * i));
    }
}

Upvotes: 21

Steinwolfe
Steinwolfe

Reputation: 81

    public static List<Color> GetGradientColors(Color start, Color end, int steps)
    {
        return GetGradientColors(start, end, steps, 0, steps - 1);
    }

    public static List<Color> GetGradientColors(Color start, Color end, int steps, int firstStep, int lastStep)
    {
        var colorList = new List<Color>();
        if (steps <= 0 || firstStep < 0 || lastStep > steps - 1)
            return colorList;

        double aStep = (end.A - start.A) / steps;
        double rStep = (end.R - start.R) / steps;
        double gStep = (end.G - start.G) / steps;
        double bStep = (end.B - start.B) / steps;

        for (int i = firstStep; i < lastStep; i++)
        {
            var a = start.A + (int)(aStep * i);
            var r = start.R + (int)(rStep * i);
            var g = start.G + (int)(gStep * i);
            var b = start.B + (int)(bStep * i);
            colorList.Add(Color.FromArgb(a, r, g, b));
        }

        return colorList;
    }

Upvotes: 8

Oliver
Oliver

Reputation: 45101

Maybe this function can help:

public IEnumerable<Color> GetGradients(Color start, Color end, int steps)
{
    Color stepper = Color.FromArgb((byte)((end.A - start.A) / (steps - 1)),
                                   (byte)((end.R - start.R) / (steps - 1)),
                                   (byte)((end.G - start.G) / (steps - 1)),
                                   (byte)((end.B - start.B) / (steps - 1)));

    for (int i = 0; i < steps; i++)
    {
        yield return Color.FromArgb(start.A + (stepper.A * i),
                                    start.R + (stepper.R * i),
                                    start.G + (stepper.G * i),
                                    start.B + (stepper.B * i));
    }
}

Upvotes: 11

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