将光频率转换为RGB？
Sha*_*ica 113 language-agnostic algorithm rgb formula approximation
有谁知道将光频率转换为RGB值的任何公式？
Ste*_*esa 40
以下是整个转换过程的详细说明:http://www.fourmilab.ch/documents/specrend/.包含源代码!
- 而Fourmilab的文章提出了一个重要的观点,即一些颜色在RGB中是不可表现的(亮橙色是一个很好的例子),因为无论我们的物理老师告诉我们什么,你都不能通过添加三种原色来"制造"任意颜色的光.我的确很好).太糟糕了,但在实践中通常不会致命. (5认同)
- 值得注意的是，RGB 颜色空间中只能准确表示所有可能的可见波长的一小部分。转换过程非常复杂和模棱两可。请参阅 https://physics.stackexchange.com/a/94446/5089 和 https://physics.stackexchange.com/a/419628/5089 (3认同)
Tar*_*arc 25
对于懒惰的人(像我一样),这里是在@ user151323的答案中找到的代码的java实现(也就是说,只是在Spectra Lab Report中找到的pascal代码的简单翻译):
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;
}
顺便说一句,这对我来说很好.
- 您的代码中似乎存在错误.如果波长为例如439.5,则函数返回黑色.我相信网站上的原始代码使用整数(我根本不知道pascal).我建议将`Wavelength <= 439`改为`Wavelength <440`. (2认同)
- 你是对的！谢谢你向我指出这一点:) 已经更正了。 (2认同)
ama*_*nov 13
大概的概念:
步骤1和2可能会有所不同.
有几种颜色匹配功能,可用作表格或分析近似值(由@Tarc和@Haochen Xie建议).如果您需要平滑的预测结果,表格是最好的.
没有单一的RGB色彩空间.可以使用多个变换矩阵和不同种类的伽马校正.
下面是我最近提出的C#代码.它在"CIE 1964标准观察者"表和sRGB矩阵+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;
}
}
400-700 nm范围的结果:
Hao*_*Xie 10
虽然这是一个老问题并且已经得到了一些很好的答案,但当我尝试在我的应用程序中实现这样的转换功能时,我对这里已经列出的算法不满意并做了我自己的研究,这给了我一些好的结果.所以我要发一个新答案.
经过一些研究后,我发现了本文,CIE XYZ颜色匹配函数的简单解析近似,并尝试在我的应用中采用引入的多瓣分段高斯拟合算法.本文仅描述了将波长转换为相应XYZ值的函数,因此我实现了一个函数,将XYZ转换为sRGB颜色空间中的RGB并合并它们.结果太棒了,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,但将它移植到较低版本的Java和其他语言并不困难.
- @Ruslan因为这个算法是CIE标准观察者的分析拟合(可以认为是"精确"模型),所以存在误差.但是从论文中,如果你看一下第7页的图1(比较(d)和(f)),这种方法提供了非常接近的近似值.特别是如果你看(f),你会发现即使在标准模型中也有蓝线.此外,纯光源的色彩感知会有所不同,因此这种误差水平可能微不足道. (3认同)
小智 7
您正在谈论从波长转换为RGB值.
看这里,可能会回答你的问题.你有一个实用程序,用源代码和一些解释这样做.
- @ Joseph Gordon - 非常不同意.考虑在空气中发出的绿色射线400nm撞击水面然后在水中传播.水的折射系数是1.33,所以水中的光线波长现在是300nm,这显然不会改变它的颜色.使光线"着色"的问题是频率,而不是波长.在相同的物质(真空,空气,水)中,频率(颜色)映射到相同的波长.在不同的媒体 - 不是. (8认同)
- 是不是几乎一样？freq = c /波长 (3认同)