Question

我想为图形图表生成随机颜色，但我尝试的是无法识别的，并且生成的颜色通常彼此相似。是否可以创造这样的颜色？

我的方法：

    public Color[] GenerateColors(int n)
    {
        Color[] colors = new Color[n];
        for (int i = 0; i < n; i++)
        {
            int R = rnd.Next(0, 250);
            int G = rnd.Next(0, 250);
            int B = rnd.Next(0, 250);

            Color color = Color.FromArgb(R, G, B);
            colors[i] = color;
        }
        return colors;
    }

Answer 1

这是一个RandomColors课程，它结合了Taw和OldBoyCoder的建议。根据您想要生成的颜色数量Generate，可以从KnownColors中选择，也可以生成随机的新颜色，使用最后一种颜色作为混合颜色。

public class RandomColors
{
    static Color lastColor = Color.Empty;

    static KnownColor[] colorValues = (KnownColor[]) Enum.GetValues(typeof(KnownColor));

    static Random rnd = new Random();
    const int MaxColor = 256;
    static RandomColors()
    {
        lastColor = Color.FromArgb(rnd.Next(MaxColor), rnd.Next(MaxColor), rnd.Next(MaxColor));
    }

    public static Color[] Generate(int n)
    {
        var colors = new Color[n];
        if (n <= colorValues.Length)
        {
            // known color suggestion from TAW
            // https://stackoverflow.com/questions/37234131/how-to-generate-randomly-colors-that-is-easily-recognizable-from-each-other#comment61999963_37234131
            var step = (colorValues.Length-1) / n;
            var colorIndex = step;
            step = step == 0 ? 1 : step; // hacky
            for(int i=0; i<n; i++ )
            {
                colors[i] = Color.FromKnownColor(colorValues[colorIndex]);
                colorIndex += step;
            }
        } else
        {
            for(int i=0; i<n; i++)
            {
                colors[i] = GetNext();
            }
        }

        return colors;
    }

    public static Color GetNext()
    {
        // use the previous value as a mix color as demonstrated by David Crow
        // https://stackoverflow.com/a/43235/578411
        Color nextColor = Color.FromArgb(
            (rnd.Next(MaxColor) + lastColor.R)/2, 
            (rnd.Next(MaxColor) + lastColor.G)/2, 
            (rnd.Next(MaxColor) + lastColor.B)/2
            );
        lastColor = nextColor;
        return nextColor;
    }
}

如何生成易于彼此识别的随机颜色？

1 个答案: