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

这在MATLAB中是微不足道的(有一个hsv命令):

cmap = hsv(number_of_colors)

其他回答

Janus的回答,但更容易读懂。我还稍微调整了配色方案,并在你可以自己修改的地方做了标记

我已经把这个片段直接粘贴到一个jupyter笔记本。

import colorsys
import itertools
from fractions import Fraction
from IPython.display import HTML as html_print

def infinite_hues():
    yield Fraction(0)
    for k in itertools.count():
        i = 2**k # zenos_dichotomy
        for j in range(1,i,2):
            yield Fraction(j,i)

def hue_to_hsvs(h: Fraction):
    # tweak values to adjust scheme
    for s in [Fraction(6,10)]:
        for v in [Fraction(6,10), Fraction(9,10)]: 
            yield (h, s, v) 

def rgb_to_css(rgb) -> str:
    uint8tuple = map(lambda y: int(y*255), rgb)
    return "rgb({},{},{})".format(*uint8tuple)

def css_to_html(css):
    return f"<text style=background-color:{css}>&nbsp;&nbsp;&nbsp;&nbsp;</text>"

def show_colors(n=33):
    hues = infinite_hues()
    hsvs = itertools.chain.from_iterable(hue_to_hsvs(hue) for hue in hues)
    rgbs = (colorsys.hsv_to_rgb(*hsv) for hsv in hsvs)
    csss = (rgb_to_css(rgb) for rgb in rgbs)
    htmls = (css_to_html(css) for css in csss)

    myhtmls = itertools.islice(htmls, n)
    display(html_print("".join(myhtmls)))

show_colors()

我认为这个简单的递归算法补充了公认的答案,以产生不同的色调值。我为hsv做了它,但也可以用于其他颜色空间。

它在循环中产生色调,在每个循环中尽可能彼此分离。

/**
 * 1st cycle: 0, 120, 240
 * 2nd cycle (+60): 60, 180, 300
 * 3th cycle (+30): 30, 150, 270, 90, 210, 330
 * 4th cycle (+15): 15, 135, 255, 75, 195, 315, 45, 165, 285, 105, 225, 345
 */
public static float recursiveHue(int n) {
    // if 3: alternates red, green, blue variations
    float firstCycle = 3;

    // First cycle
    if (n < firstCycle) {
        return n * 360f / firstCycle;
    }
    // Each cycle has as much values as all previous cycles summed (powers of 2)
    else {
        // floor of log base 2
        int numCycles = (int)Math.floor(Math.log(n / firstCycle) / Math.log(2));
        // divDown stores the larger power of 2 that is still lower than n
        int divDown = (int)(firstCycle * Math.pow(2, numCycles));
        // same hues than previous cycle, but summing an offset (half than previous cycle)
        return recursiveHue(n % divDown) + 180f / divDown;
    }
}

我在这里找不到这种算法。我希望这对你有所帮助,这是我在这里的第一篇文章。

我会尽量把饱和度和亮度调到最大,只关注色调。在我看来,H可以从0到255,然后绕圈。现在如果你想要两种对比色,你可以取这个环的对边,即0和128。如果你想要4种颜色,你需要取一些以圆周长度256的1/4为间隔的颜色,即0,64,128,192。当然,正如其他人建议的那样,当你需要N种颜色时,你可以用256/N将它们分开。

我想补充的是,用二进制数的反向表示来形成这个序列。看看这个:

0 = 00000000  after reversal is 00000000 = 0
1 = 00000001  after reversal is 10000000 = 128
2 = 00000010  after reversal is 01000000 = 64
3 = 00000011  after reversal is 11000000 = 192

... 这样,如果你需要N种不同的颜色,你只需要取前N个数字,把它们倒过来,你就能得到尽可能多的距离点(因为N是2的幂),同时保持序列的每个前缀都有很大不同。

在我的用例中,这是一个重要的目标,因为我有一个图表,其中颜色是根据这种颜色所覆盖的区域进行排序的。我希望图表中最大的区域具有较大的对比度,并且我对一些小区域使用与前十名相似的颜色也没有问题,因为读者通过观察区域就可以很明显地看出哪个是哪个。

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))

为了子孙后代,我在这里添加了Python中公认的答案。

import numpy as np
import colorsys

def _get_colors(num_colors):
    colors=[]
    for i in np.arange(0., 360., 360. / num_colors):
        hue = i/360.
        lightness = (50 + np.random.rand() * 10)/100.
        saturation = (90 + np.random.rand() * 10)/100.
        colors.append(colorsys.hls_to_rgb(hue, lightness, saturation))
    return colors