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);
}
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)]
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。
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
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。
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)]
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);
}
