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);
}

当前回答

如果N足够大,你会得到一些相似的颜色。世界上只有这么多。

为什么不把它们均匀地分布在光谱中,像这样:

IEnumerable<Color> CreateUniqueColors(int nColors)
{
    int subdivision = (int)Math.Floor(Math.Pow(nColors, 1/3d));
    for(int r = 0; r < 255; r += subdivision)
        for(int g = 0; g < 255; g += subdivision)
            for(int b = 0; b < 255; b += subdivision)
                yield return Color.FromArgb(r, g, b);
}

如果您想混合序列,以便相似的颜色不在彼此旁边,您可能会打乱结果列表。

是我想得不够周全吗?

其他回答

这里有一个解决你的“独特”问题的解决方案,这完全是夸大的:

创建一个单位球体,并在其上放置带有排斥电荷的点。运行一个粒子系统,直到它们不再移动(或者delta“足够小”)。在这一点上,每个点之间的距离都尽可能远。将(x, y, z)转换为rgb。

我提到它是因为对于某些类型的问题,这种类型的解决方案比暴力解决方案更好。

我一开始看到这种方法是用来镶嵌球面的。

同样,遍历HSL空间或RGB空间的最明显的解决方案可能工作得很好。

如果N足够大,你会得到一些相似的颜色。世界上只有这么多。

为什么不把它们均匀地分布在光谱中,像这样:

IEnumerable<Color> CreateUniqueColors(int nColors)
{
    int subdivision = (int)Math.Floor(Math.Pow(nColors, 1/3d));
    for(int r = 0; r < 255; r += subdivision)
        for(int g = 0; g < 255; g += subdivision)
            for(int b = 0; b < 255; b += subdivision)
                yield return Color.FromArgb(r, g, b);
}

如果您想混合序列,以便相似的颜色不在彼此旁边,您可能会打乱结果列表。

是我想得不够周全吗?

这个OpenCV函数使用HSV颜色模型在0<=H<=360º周围生成n个均匀分布的颜色,最大S=1.0, V=1.0。函数在bgr_mat中输出BGR颜色:

void distributed_colors (int n, cv::Mat_<cv::Vec3f> & bgr_mat) {
  cv::Mat_<cv::Vec3f> hsv_mat(n,CV_32F,cv::Vec3f(0.0,1.0,1.0));
  double step = 360.0/n;
  double h= 0.0;
  cv::Vec3f value;
  for (int i=0;i<n;i++,h+=step) {
    value = hsv_mat.at<cv::Vec3f>(i);
    hsv_mat.at<cv::Vec3f>(i)[0] = h;
  }
  cv::cvtColor(hsv_mat, bgr_mat, CV_HSV2BGR);
  bgr_mat *= 255;
}

这产生了与Janus Troelsen的溶液相同的颜色。但是它使用的不是生成器,而是开始/停止语义。它也是完全向量化的。

import numpy as np
import numpy.typing as npt
import matplotlib.colors

def distinct_colors(start: int=0, stop: int=20) -> npt.NDArray[np.float64]:
    """Returns an array of distinct RGB colors, from an infinite sequence of colors
    """
    if stop <= start: # empty interval; return empty array
        return np.array([], dtype=np.float64)
    sat_values = [6/10]         # other tones could be added
    val_values = [8/10, 5/10]   # other tones could be added
    colors_per_hue_value = len(sat_values) * len(val_values)
    # Get the start and stop indices within the hue value stream that are needed
    # to achieve the requested range
    hstart = start // colors_per_hue_value
    hstop = (stop+colors_per_hue_value-1) // colors_per_hue_value
    # Zero will cause a singularity in the caluculation, so we will add the zero
    # afterwards
    prepend_zero = hstart==0 

    # Sequence (if hstart=1): 1,2,...,hstop-1
    i = np.arange(1 if prepend_zero else hstart, hstop) 
    # The following yields (if hstart is 1): 1/2,  1/4, 3/4,  1/8, 3/8, 5/8, 7/8,  
    # 1/16, 3/16, ... 
    hue_values = (2*i+1) / np.power(2,np.floor(np.log2(i*2))) - 1
    
    if prepend_zero:
        hue_values = np.concatenate(([0], hue_values))

    # Make all combinations of h, s and v values, as if done by a nested loop
    # in that order
    hsv = np.array(np.meshgrid(hue_values, sat_values, val_values, indexing='ij')
                    ).reshape((3,-1)).transpose()

    # Select the requested range (only the necessary values were computed but we
    # need to adjust the indices since start & stop are not necessarily multiples
    # of colors_per_hue_value)
    hsv = hsv[start % colors_per_hue_value : 
                start % colors_per_hue_value + stop - start]
    # Use the matplotlib vectorized function to convert hsv to rgb
    return matplotlib.colors.hsv_to_rgb(hsv)

样品:

from matplotlib.colors import ListedColormap
ListedColormap(distinct_colors(stop=20))

ListedColormap(distinct_colors(start=30, stop=50))

对于Python用户来说,seaborn非常简洁:

>>> import seaborn as sns
>>> sns.color_palette(n_colors=4)

它返回RGB元组列表:

[(0.12156862745098039, 0.4666666666666667, 0.7058823529411765),
(1.0, 0.4980392156862745, 0.054901960784313725),
(0.17254901960784313, 0.6274509803921569, 0.17254901960784313),
(0.8392156862745098, 0.15294117647058825, 0.1568627450980392)]