将光频转换为RGB？

有谁知道任何公式将光频转换为RGB值？

Answer 1

以下是整个转换过程的详细说明： http ： //www.fourmilab.ch/documents/specrend/ 。 包括源代码！

Answer 2

对于懒惰的人（像我），这里是在@ user151323的答案（即，只是从光谱实验室报告中发现的pascal代码简单的翻译）中find的代码的Java实现：

static private double Gamma = 0.80; static private double IntensityMax = 255; /** Taken from Earl F. Glynn's web page: * <a href="http://www.efg2.com/Lab/ScienceAndEngineering/Spectra.htm">Spectra Lab Report</a> * */ public static int[] waveLengthToRGB(double Wavelength){ double factor; double Red,Green,Blue; if((Wavelength >= 380) && (Wavelength<440)){ Red = -(Wavelength - 440) / (440 - 380); Green = 0.0; Blue = 1.0; }else if((Wavelength >= 440) && (Wavelength<490)){ Red = 0.0; Green = (Wavelength - 440) / (490 - 440); Blue = 1.0; }else if((Wavelength >= 490) && (Wavelength<510)){ Red = 0.0; Green = 1.0; Blue = -(Wavelength - 510) / (510 - 490); }else if((Wavelength >= 510) && (Wavelength<580)){ Red = (Wavelength - 510) / (580 - 510); Green = 1.0; Blue = 0.0; }else if((Wavelength >= 580) && (Wavelength<645)){ Red = 1.0; Green = -(Wavelength - 645) / (645 - 580); Blue = 0.0; }else if((Wavelength >= 645) && (Wavelength<781)){ Red = 1.0; Green = 0.0; Blue = 0.0; }else{ Red = 0.0; Green = 0.0; Blue = 0.0; }; // Let the intensity fall off near the vision limits if((Wavelength >= 380) && (Wavelength<420)){ factor = 0.3 + 0.7*(Wavelength - 380) / (420 - 380); }else if((Wavelength >= 420) && (Wavelength<701)){ factor = 1.0; }else if((Wavelength >= 701) && (Wavelength<781)){ factor = 0.3 + 0.7*(780 - Wavelength) / (780 - 700); }else{ factor = 0.0; }; int[] rgb = new int[3]; // Don't want 0^x = 1 for x <> 0 rgb[0] = Red==0.0 ? 0 : (int) Math.round(IntensityMax * Math.pow(Red * factor, Gamma)); rgb[1] = Green==0.0 ? 0 : (int) Math.round(IntensityMax * Math.pow(Green * factor, Gamma)); rgb[2] = Blue==0.0 ? 0 : (int) Math.round(IntensityMax * Math.pow(Blue * factor, Gamma)); return rgb; } 

顺便说一下，这对我来说工作得很好。

Answer 3

你正在谈论从波长转换为RGB值。

看这里，可能会回答你的问题。 你有一个这样的源代码以及一些解释的工具。

WaveLengthToRGB

Answer 4

虽然这是一个古老的问题，已经得到了一些很好的答案，当我试图在我的应用程序中实现这种转换function时，我不满意这里列出的algorithm，并做了我自己的研究，这给了我一些好的结果。 所以我要发布一个新的答案。

经过一些研究，我遇到了本文， 对CIE XYZ颜色匹配函数的简单分析近似 ，并试图在我的应用中采用引入的多叶片分段高斯拟合algorithm。 本文仅描述了将波长转换为相应XYZ值的function，因此我实现了将srGB颜色空间中的XYZ转换为RGB的function并将它们组合在一起。 结果是太棒了，值得分享：

 /** * Convert a wavelength in the visible light spectrum to a RGB color value that is suitable to be displayed on a * monitor * * @param wavelength wavelength in nm * @return RGB color encoded in int. each color is represented with 8 bits and has a layout of * 00000000RRRRRRRRGGGGGGGGBBBBBBBB where MSB is at the leftmost */ public static int wavelengthToRGB(double wavelength){ double[] xyz = cie1931WavelengthToXYZFit(wavelength); double[] rgb = srgbXYZ2RGB(xyz); int c = 0; c |= (((int) (rgb[0] * 0xFF)) & 0xFF) << 16; c |= (((int) (rgb[1] * 0xFF)) & 0xFF) << 8; c |= (((int) (rgb[2] * 0xFF)) & 0xFF) << 0; return c; } /** * Convert XYZ to RGB in the sRGB color space * <p> * The conversion matrix and color component transfer function is taken from http://www.color.org/srgb.pdf, which * follows the International Electrotechnical Commission standard IEC 61966-2-1 "Multimedia systems and equipment - * Colour measurement and management - Part 2-1: Colour management - Default RGB colour space - sRGB" * * @param xyz XYZ values in a double array in the order of X, Y, Z. each value in the range of [0.0, 1.0] * @return RGB values in a double array, in the order of R, G, B. each value in the range of [0.0, 1.0] */ public static double[] srgbXYZ2RGB(double[] xyz) { double x = xyz[0]; double y = xyz[1]; double z = xyz[2]; double rl = 3.2406255 * x + -1.537208 * y + -0.4986286 * z; double gl = -0.9689307 * x + 1.8757561 * y + 0.0415175 * z; double bl = 0.0557101 * x + -0.2040211 * y + 1.0569959 * z; return new double[] { srgbXYZ2RGBPostprocess(rl), srgbXYZ2RGBPostprocess(gl), srgbXYZ2RGBPostprocess(bl) }; } /** * helper function for {@link #srgbXYZ2RGB(double[])} */ private static double srgbXYZ2RGBPostprocess(double c) { // clip if c is out of range c = c > 1 ? 1 : (c < 0 ? 0 : c); // apply the color component transfer function c = c <= 0.0031308 ? c * 12.92 : 1.055 * Math.pow(c, 1. / 2.4) - 0.055; return c; } /** * A multi-lobe, piecewise Gaussian fit of CIE 1931 XYZ Color Matching Functions by Wyman el al. from Nvidia. The * code here is adopted from the Listing 1 of the paper authored by Wyman et al. * <p> * Reference: Chris Wyman, Peter-Pike Sloan, and Peter Shirley, Simple Analytic Approximations to the CIE XYZ Color * Matching Functions, Journal of Computer Graphics Techniques (JCGT), vol. 2, no. 2, 1-11, 2013. * * @param wavelength wavelength in nm * @return XYZ in a double array in the order of X, Y, Z. each value in the range of [0.0, 1.0] */ public static double[] cie1931WavelengthToXYZFit(double wavelength) { double wave = wavelength; double x; { double t1 = (wave - 442.0) * ((wave < 442.0) ? 0.0624 : 0.0374); double t2 = (wave - 599.8) * ((wave < 599.8) ? 0.0264 : 0.0323); double t3 = (wave - 501.1) * ((wave < 501.1) ? 0.0490 : 0.0382); x = 0.362 * Math.exp(-0.5 * t1 * t1) + 1.056 * Math.exp(-0.5 * t2 * t2) - 0.065 * Math.exp(-0.5 * t3 * t3); } double y; { double t1 = (wave - 568.8) * ((wave < 568.8) ? 0.0213 : 0.0247); double t2 = (wave - 530.9) * ((wave < 530.9) ? 0.0613 : 0.0322); y = 0.821 * Math.exp(-0.5 * t1 * t1) + 0.286 * Math.exp(-0.5 * t2 * t2); } double z; { double t1 = (wave - 437.0) * ((wave < 437.0) ? 0.0845 : 0.0278); double t2 = (wave - 459.0) * ((wave < 459.0) ? 0.0385 : 0.0725); z = 1.217 * Math.exp(-0.5 * t1 * t1) + 0.681 * Math.exp(-0.5 * t2 * t2); } return new double[] { x, y, z }; } 

我的代码是用Java 8编写的，但不应该很难把它移植到较低版本的Java和其他语言。

Answer 5

大概的概念：

1. 使用CEI颜色匹配function将波长转换为XYZ颜色 。
2. 将XYZ转换为RGB
3. 将组件剪切为[0..1]并乘以255以适应无符号的字节范围。

步骤1和2可能会有所不同。

有几种颜色匹配function，可作为表格或作为parsing近似（由@Tarc和@Hocochen Xiebuild议）。 如果你需要一个平稳的预测结果，表格是最好的。

没有单一的RGB色彩空间。 可以使用多个变换matrix和不同types的伽马校正。

下面是我最近提出的C＃代码。 它在“CIE 1964标准观察者”表和sRGBmatrix+伽马校正上使用线性插值。

 static class RgbCalculator { const int LEN_MIN = 380, LEN_MAX = 780, LEN_STEP = 5; static readonly double[] X = { 0.000160, 0.000662, 0.002362, 0.007242, 0.019110, 0.043400, 0.084736, 0.140638, 0.204492, 0.264737, 0.314679, 0.357719, 0.383734, 0.386726, 0.370702, 0.342957, 0.302273, 0.254085, 0.195618, 0.132349, 0.080507, 0.041072, 0.016172, 0.005132, 0.003816, 0.015444, 0.037465, 0.071358, 0.117749, 0.172953, 0.236491, 0.304213, 0.376772, 0.451584, 0.529826, 0.616053, 0.705224, 0.793832, 0.878655, 0.951162, 1.014160, 1.074300, 1.118520, 1.134300, 1.123990, 1.089100, 1.030480, 0.950740, 0.856297, 0.754930, 0.647467, 0.535110, 0.431567, 0.343690, 0.268329, 0.204300, 0.152568, 0.112210, 0.081261, 0.057930, 0.040851, 0.028623, 0.019941, 0.013842, 0.009577, 0.006605, 0.004553, 0.003145, 0.002175, 0.001506, 0.001045, 0.000727, 0.000508, 0.000356, 0.000251, 0.000178, 0.000126, 0.000090, 0.000065, 0.000046, 0.000033 }, Y = { 0.000017, 0.000072, 0.000253, 0.000769, 0.002004, 0.004509, 0.008756, 0.014456, 0.021391, 0.029497, 0.038676, 0.049602, 0.062077, 0.074704, 0.089456, 0.106256, 0.128201, 0.152761, 0.185190, 0.219940, 0.253589, 0.297665, 0.339133, 0.395379, 0.460777, 0.531360, 0.606741, 0.685660, 0.761757, 0.823330, 0.875211, 0.923810, 0.961988, 0.982200, 0.991761, 0.999110, 0.997340, 0.982380, 0.955552, 0.915175, 0.868934, 0.825623, 0.777405, 0.720353, 0.658341, 0.593878, 0.527963, 0.461834, 0.398057, 0.339554, 0.283493, 0.228254, 0.179828, 0.140211, 0.107633, 0.081187, 0.060281, 0.044096, 0.031800, 0.022602, 0.015905, 0.011130, 0.007749, 0.005375, 0.003718, 0.002565, 0.001768, 0.001222, 0.000846, 0.000586, 0.000407, 0.000284, 0.000199, 0.000140, 0.000098, 0.000070, 0.000050, 0.000036, 0.000025, 0.000018, 0.000013 }, Z = { 0.000705, 0.002928, 0.010482, 0.032344, 0.086011, 0.197120, 0.389366, 0.656760, 0.972542, 1.282500, 1.553480, 1.798500, 1.967280, 2.027300, 1.994800, 1.900700, 1.745370, 1.554900, 1.317560, 1.030200, 0.772125, 0.570060, 0.415254, 0.302356, 0.218502, 0.159249, 0.112044, 0.082248, 0.060709, 0.043050, 0.030451, 0.020584, 0.013676, 0.007918, 0.003988, 0.001091, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000, 0.000000 }; static readonly double[] MATRIX_SRGB_D65 = { 3.2404542, -1.5371385, -0.4985314, -0.9692660, 1.8760108, 0.0415560, 0.0556434, -0.2040259, 1.0572252 }; public static byte[] Calc(double len) { if(len < LEN_MIN || len > LEN_MAX) return new byte[3]; len -= LEN_MIN; var index = (int)Math.Floor(len / LEN_STEP); var offset = len - LEN_STEP * index; var x = Interpolate(X, index, offset); var y = Interpolate(Y, index, offset); var z = Interpolate(Z, index, offset); var m = MATRIX_SRGB_D65; var r = m[0] * x + m[1] * y + m[2] * z; var g = m[3] * x + m[4] * y + m[5] * z; var b = m[6] * x + m[7] * y + m[8] * z; r = Clip(GammaCorrect_sRGB(r)); g = Clip(GammaCorrect_sRGB(g)); b = Clip(GammaCorrect_sRGB(b)); return new[] { (byte)(255 * r), (byte)(255 * g), (byte)(255 * b) }; } static double Interpolate(double[] values, int index, double offset) { if(offset == 0) return values[index]; var x0 = index * LEN_STEP; var x1 = x0 + LEN_STEP; var y0 = values[index]; var y1 = values[1 + index]; return y0 + offset * (y1 - y0) / (x1 - x0); } static double GammaCorrect_sRGB(double c) { if(c <= 0.0031308) return 12.92 * c; var a = 0.055; return (1 + a) * Math.Pow(c, 1 / 2.4) - a; } static double Clip(double c) { if(c < 0) return 0; if(c > 1) return 1; return c; } } 

400-700nm范围的结果：

Answer 6

我想我也可以用正式的答案来跟进我的评论。 最好的select是使用HSV色彩空间 – 虽然色调代表波长，但它不是一对一的比较。

Answer 7

我做了一个已知的色调值和频率的线性拟合（褪去红色和紫色，因为它们的频率值延伸到目前为止有些偏差），我得到了一个粗略的转换公式。

它就像
频率（单位THz）= 474 +（3/4）（色调angular度（度））

我试图环顾四周，看看有没有人提出这个等式，但是到2010年5月我还没有发现任何东西。