在C#中生成颜色渐变

Question

我的问题类似于这里的问题,除了我正在使用C#.

我有两种颜色,我有一个预定义的步骤.如何检索Color两者之间渐变的s 列表？

这是我尝试过的方法,但没有用:

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));
}

如果你尝试上面的代码,你会发现逐渐增加argb并不对应于颜色的视觉逐渐增加.

有什么想法吗？

Answer 1

您必须提取R,G,B分量并分别对它们中的每一个执行相同的线性插值,然后重新组合.

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));
}

Answer 2

奥利弗的答案非常接近......但在我的情况下,我的一些步进数字必须是负数.当将步进器值转换为Color结构时,我的值从负值变为更高的值,例如-1变为类似于254的值.我单独设置我的步长值来解决这个问题.

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));
    }
}

Answer 3

也许这个功能可以帮助:

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));
    }
}

Answer 4

    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;
    }

Answer 5

使用double而不是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));

和：

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));