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);
}
我为R写了一个名为qualpalr的包，它是专门为此目的设计的。我建议你看看小插图，看看它是如何工作的，但我会尽量总结要点。
qualpalr在HSL颜色空间(前面在这个线程中描述过)中获取一个颜色规范，将其投射到DIN99d颜色空间(感知上是均匀的)，并找到使它们之间的最小距离最大化的n。
# Create a palette of 4 colors of hues from 0 to 360, saturations between
# 0.1 and 0.5, and lightness from 0.6 to 0.85
pal <- qualpal(n = 4, list(h = c(0, 360), s = c(0.1, 0.5), l = c(0.6, 0.85)))
# Look at the colors in hex format
pal$hex
#> [1] "#6F75CE" "#CC6B76" "#CAC16A" "#76D0D0"
# Create a palette using one of the predefined color subspaces
pal2 <- qualpal(n = 4, colorspace = "pretty")
# Distance matrix of the DIN99d color differences
pal2$de_DIN99d
#> #69A3CC #6ECC6E #CA6BC4
#> 6ECC6E 22
#> CA6BC4 21 30
#> CD976B 24 21 21
plot(pal2)
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。
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}> </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()
