如何自动生成N个“distinct"颜色? [英] How to automatically generate N "distinct" colors?

问题描述

我写了下面两个方法来自动选择N个不同的颜色.它的工作原理是在 RGB 立方体上定义分段线性函数.这样做的好处是，如果您想要的话，您还可以获得渐进式比例，但是当 N 变大时，颜色可能会开始看起来相似.我还可以想象将RGB立方体均匀地细分为一个点阵，然后绘制点.有谁知道其他方法吗?我排除了定义一个列表然后循环遍历它的可能性.我还应该说，我通常不在乎它们是否发生冲突或不好看，它们只需要在视觉上与众不同.

I wrote the two methods below to automatically select N distinct colors. It works by defining a piecewise linear function on the RGB cube. The benefit of this is you can also get a progressive scale if that's what you want, but when N gets large the colors can start to look similar. I can also imagine evenly subdividing the RGB cube into a lattice and then drawing points. Does anyone know any other methods? I'm ruling out defining a list and then just cycling through it. I should also say I don't generally care if they clash or don't look nice, they just have to be visually distinct.

public static List<Color> pick(int num) {
    List<Color> colors = new ArrayList<Color>();
    if (num < 2)
        return colors;
    float dx = 1.0f / (float) (num - 1);
    for (int i = 0; i < num; i++) {
        colors.add(get(i * dx));
    }
    return colors;
}

public static Color get(float x) {
    float r = 0.0f;
    float g = 0.0f;
    float b = 1.0f;
    if (x >= 0.0f && x < 0.2f) {
        x = x / 0.2f;
        r = 0.0f;
        g = x;
        b = 1.0f;
    } else if (x >= 0.2f && x < 0.4f) {
        x = (x - 0.2f) / 0.2f;
        r = 0.0f;
        g = 1.0f;
        b = 1.0f - x;
    } else if (x >= 0.4f && x < 0.6f) {
        x = (x - 0.4f) / 0.2f;
        r = x;
        g = 1.0f;
        b = 0.0f;
    } else if (x >= 0.6f && x < 0.8f) {
        x = (x - 0.6f) / 0.2f;
        r = 1.0f;
        g = 1.0f - x;
        b = 0.0f;
    } else if (x >= 0.8f && x <= 1.0f) {
        x = (x - 0.8f) / 0.2f;
        r = 1.0f;
        g = 0.0f;
        b = x;
    }
    return new Color(r, g, b);
}

推荐答案

您可以使用 HSL 颜色模型 来创建你的颜色.

You can use the HSL color model to create your colors.

如果您只想要不同的色调(可能)，以及亮度或饱和度的细微变化，您可以像这样分配色调:

If all you want is differing hues (likely), and slight variations on lightness or saturation, you can distribute the hues like so:

// assumes hue [0, 360), saturation [0, 100), lightness [0, 100)

for(i = 0; i < 360; i += 360 / num_colors) {
    HSLColor c;
    c.hue = i;
    c.saturation = 90 + randf() * 10;
    c.lightness = 50 + randf() * 10;

    addColor(c);
}

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