寻找N种不同的RGB颜色

Question

我试图以图形方式显示N行的图形,我正在尝试根据我有多少行来找到一种动态分配不同颜色的方法.RGB中的值范围为0到1.由于背景为白色,因此无法使用白色.我发现N <7很容易:

r=(h&0x4)/4;
g=(h&0x2)/2;
b=h&0x1;

这给了我黑色,蓝色,绿色,青色,红色,洋红色,黄色.但之后它将使用白色然后循环.有人知道为索引分配RGB值的好方法吗？我也有一个不透明度的值.

Answer 1

我这样做的首选方法是n沿着色轮找到均匀间隔的点.

我们代表色轮为0和360之间的值的范围.因此,我们将使用值360 / n * 0,360 / n * 1,..., 360 / n * (n - 1).在这一过程中,我们已经定义了色相我们每一个颜色的.我们可以通过将饱和度设置为1并将亮度设置为1来将这些颜色中的每一种描述为色调饱和度值(HSV)颜色.

(饱和度越高意味着颜色越"丰富";饱和度越低意味着颜色越接近灰色;亮度越高意味着颜色越"亮";亮度越低意味着颜色越"暗".

现在,一个简单的计算给出了每种颜色的RGB值.

http://en.wikipedia.org/wiki/HSL_and_HSV#Conversion_from_HSV_to_RGB

请注意,给出的方程式可以简化:

• p = v*(1-s)= 1*(1 - 1)= 1*0 = 0
• q = v*(1-f*s)= 1*(1-f*1)= 1-f
• t = v*(1 - (1 - f)*s)= 1*(1 - (1 - f)*1)= 1 - (1 - f)= 1 - 1 + f = f

Python中的伪代码实现

注意:这是一个非常低效的实现.在Python中给出这个例子的要点基本上是我可以给出可执行的伪代码.

import math

def uniquecolors(n):
    """Compute a list of distinct colors, each of which is represented as an RGB 3-tuple."""
    hues = []
    # i is in the range 0, 1, ..., n - 1
    for i in range(n):
        hues.append(360.0 / i)

    hs = []
    for hue in hues:
        h = math.floor(hue / 60) % 6
        hs.append(h)

    fs = []
    for hue in hues:
        f = hue / 60 - math.floor(hue / 60)
        fs.append(f)

    rgbcolors = []
    for h, f in zip(hs, fs):
        v = 1
        p = 0
        q = 1 - f
        t = f
        if h == 0:
            color = v, t, p
        elif h == 1:
            color = q, v, p
        elif h == 2:
            color = p, v, t
        elif h == 3:
            color = p, q, v
        elif h == 4:
            color = t, p, v
        elif h == 5:
            color = v, p, q
        rgbcolors.append(color)

    return rgbcolors

Python中的简明实现

import math

v = 1.0
s = 1.0
p = 0.0
def rgbcolor(h, f):
    """Convert a color specified by h-value and f-value to an RGB
    three-tuple."""
    # q = 1 - f
    # t = f
    if h == 0:
        return v, f, p
    elif h == 1:
        return 1 - f, v, p
    elif h == 2:
        return p, v, f
    elif h == 3:
        return p, 1 - f, v
    elif h == 4:
        return f, p, v
    elif h == 5:
        return v, p, 1 - f

def uniquecolors(n):
    """Compute a list of distinct colors, ecah of which is
    represented as an RGB three-tuple"""
    hues = (360.0 / n * i for i in range(n))
    hs = (math.floor(hue / 60) % 6 for hue in hues)
    fs = (hue / 60 - math.floor(hue / 60) for hue in hues)
    return [rgbcolor(h, f) for h, f in zip(hs, fs)]