查找N个不同的RGB颜色

本文介绍了查找N个不同的RGB颜色的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 我试图以图形方式显示N行的图形，我试图找到一种方法来动态分配不同的颜色，基于我有多少行。 RGB中的值范围从0到1.我不能使用白色，因为背景是白色的。我发现N I'm trying to graphically display a graph of N lines and I'm trying to find a way to dynamically assign distinct colors based on how many lines I have. The values in RGB range from 0 to 1. I can't use white because the background is white. I found it easy for N < 7:r=(h&0x4)/4;g=(h&0x2)/2;b=h&0x1;这给我黑色，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。有人知道一个好的方法来为索引分配RGB值吗？我还有一个不透明度值来玩。This gives me black, blue, green, cyan, red, magenta, yellow. But after that it will use white and then loop. Does anybody know a good way to assign RGB values for an index? I also have an opacity value to play with.推荐答案我首选的方法是找到 My preferred method for doing this is to find n evenly-spaced points along the colour wheel.我们将色轮表示为介于0和360之间的值范围。因此，我们将使用的值是 360 / n * 0 ， 360 / n * 1 ，...， 360 / n *（n-1）。为此，我们定义了每种颜色的 hue 。我们可以通过将饱和度设置为1和亮度设置为1来将这些颜色描述为色相 - 饱和度值（HSV）颜色。We represent the colour wheel as a range of values between 0 and 360. Thus, the values we will use are 360 / n * 0, 360 / n * 1, ..., 360 / n * (n - 1). In doing this, we've defined the hue of each of our colours. We can describe each of these colours as Hue-Saturation-Value (HSV) colours by setting saturation to 1 and lightness to 1.更丰富;较低的饱和度意味着颜色更接近灰色;较高的亮度意味着颜色是较亮的;较低的亮度意味着颜色是较暗的。）(A higher saturation means the colour is more "rich"; a lower saturation means the colour is closer to gray. A higher lightness means the colour is "brighter"; a lower lightness means the colour is "darker".)现在，一个简单的计算给出了每种颜色的RGB值。Now, a simple calculation gives us the RGB values of each of these colours. http://en.wikipedia.org/wiki/HSL_and_HSV#Conversion_from_HSV_to_RGB 请注意，可以简化给出的方程：Note that the equations given can be simplified: p = v *（1 - s）= 1 * 1）= 1 * 0 = 0 q = v *（1 - f * s） f * 1）= 1 - f t = v *（1 - （1 - f）* s）= 1 *（1-（1-f）* 1）= 1-（1-f）= 1-1 + f = f p = v * (1 - s) = 1 * (1 - 1) = 1 * 0 = 0q = v * (1 - f * s) = 1 * (1 - f * 1) = 1 - ft = v * (1 - (1 - f) * s) = 1 * (1 - (1 - f) * 1) = 1 - (1 - f) = 1 - 1 + f = f注意：这是一个非常低效的实现。在Python中给出这个例子的本质是我可以给出可执行的伪代码。Note: This is intentionally a horribly inefficient implementation. The point of giving this example in Python is essentially so I can give executable pseudocode.import mathdef 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 在Python中简明实现 Concise Implementation in Pythonimport mathv = 1.0s = 1.0p = 0.0def 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 - fdef 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)] 这篇关于查找N个不同的RGB颜色的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！