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

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。

其他回答

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。

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?

您可以使用HSL颜色模型来创建颜色。

如果你想要的只是不同的色调(可能),以及亮度或饱和度的轻微变化,你可以像这样分配色调:

// 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);
}

就像Uri Cohen的答案,但它是一个生成器。首先要把颜色分开。确定的。

样品,左边颜色先:

#!/usr/bin/env python3
from typing import Iterable, Tuple
import colorsys
import itertools
from fractions import Fraction
from pprint import pprint

def zenos_dichotomy() -> Iterable[Fraction]:
    """
    http://en.wikipedia.org/wiki/1/2_%2B_1/4_%2B_1/8_%2B_1/16_%2B_%C2%B7_%C2%B7_%C2%B7
    """
    for k in itertools.count():
        yield Fraction(1,2**k)

def fracs() -> Iterable[Fraction]:
    """
    [Fraction(0, 1), Fraction(1, 2), Fraction(1, 4), Fraction(3, 4), Fraction(1, 8), Fraction(3, 8), Fraction(5, 8), Fraction(7, 8), Fraction(1, 16), Fraction(3, 16), ...]
    [0.0, 0.5, 0.25, 0.75, 0.125, 0.375, 0.625, 0.875, 0.0625, 0.1875, ...]
    """
    yield Fraction(0)
    for k in zenos_dichotomy():
        i = k.denominator # [1,2,4,8,16,...]
        for j in range(1,i,2):
            yield Fraction(j,i)

# can be used for the v in hsv to map linear values 0..1 to something that looks equidistant
# bias = lambda x: (math.sqrt(x/3)/Fraction(2,3)+Fraction(1,3))/Fraction(6,5)

HSVTuple = Tuple[Fraction, Fraction, Fraction]
RGBTuple = Tuple[float, float, float]

def hue_to_tones(h: Fraction) -> Iterable[HSVTuple]:
    for s in [Fraction(6,10)]: # optionally use range
        for v in [Fraction(8,10),Fraction(5,10)]: # could use range too
            yield (h, s, v) # use bias for v here if you use range

def hsv_to_rgb(x: HSVTuple) -> RGBTuple:
    return colorsys.hsv_to_rgb(*map(float, x))

flatten = itertools.chain.from_iterable

def hsvs() -> Iterable[HSVTuple]:
    return flatten(map(hue_to_tones, fracs()))

def rgbs() -> Iterable[RGBTuple]:
    return map(hsv_to_rgb, hsvs())

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

def css_colors() -> Iterable[str]:
    return map(rgb_to_css, rgbs())

if __name__ == "__main__":
    # sample 100 colors in css format
    sample_colors = list(itertools.islice(css_colors(), 100))
    pprint(sample_colors)

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()