有谁知道将光频率转换为 RGB 值的任何公式？
问问题
69413 次
9 回答
49
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！
于 2009-09-24T15:47:33.740 回答
29
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，只是从Spectra Lab Report中找到的 pascal 代码的简单翻译）：
static private final double Gamma = 0.80;
static private final 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;
}
于 2013-02-17T02:46:49.130 回答
16
大概的概念：
- 使用CEI 颜色匹配功能将波长转换为XYZ 颜色。
- 将 XYZ 转换为 RGB
- 将组件剪辑到 [0..1] 并乘以 255 以适应无符号字节范围。
步骤 1 和 2 可能会有所不同。
有几种颜色匹配函数，可作为表格或解析近似值（@Tarc 和 @Haochen Xie 建议）。如果您需要平滑精确的结果，最好使用表格。
没有单一的 RGB 颜色空间。可以使用多个变换矩阵和不同种类的伽马校正。
下面是我最近想出的 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：
于 2016-09-12T08:44:03.100 回答
10
虽然这是一个老问题，并且已经得到了一些好的答案，但当我尝试在我的应用程序中实现这种转换功能时，我对这里已经列出的算法并不满意，我自己进行了研究，这给了我一些很好的结果。所以我要发布一个新的答案。
经过一些研究，我发现了这篇论文，Simple Analytic Approximations to the CIE XYZ Color Matching Functions，并尝试在我的应用程序中采用引入的多叶分段高斯拟合算法。论文只描述了将波长转换为对应的XYZ 值的函数，因此我在 sRGB 颜色空间中实现了将 XYZ 转换为 RGB 的函数并将它们组合起来。结果非常棒，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，但将其移植到较低版本的 Java 和其他语言应该不难。
于 2016-01-03T21:22:53.070 回答
7
于 2009-09-24T15:45:03.260 回答
3
我想我不妨用一个正式的答案来跟进我的评论。最好的选择是使用HSV 颜色空间- 虽然色调代表波长，但它不是一对一的比较。
于 2009-09-24T15:53:10.500 回答
3
我对已知的色调值和频率进行了线性拟合（去掉了红色和紫色，因为它们在频率值上延伸得太远以至于它们有点歪斜），我得到了一个粗略的转换方程。
它就像
频率（THz）=474+（3/4）（色相角（度））
我试图环顾四周，看看是否有人提出了这个等式，但截至 2010 年 5 月我还没有找到任何东西。
于 2010-05-13T23:58:18.023 回答
2
方法一
这是对@haochen-xie 的 C++11 版本进行了一些清理和测试。我还添加了一个函数，将值 0 转换为 1 为可用于此方法的可见光谱波长。您可以将其放在一个头文件中并在没有任何依赖关系的情况下使用它。此版本将保留在这里。
#ifndef common_utils_OnlineStats_hpp
#define common_utils_OnlineStats_hpp
namespace common_utils {
class ColorUtils {
public:
static void valToRGB(double val0To1, unsigned char& r, unsigned char& g, unsigned char& b)
{
//actual visible spectrum is 375 to 725 but outside of 400-700 things become too dark
wavelengthToRGB(val0To1 * (700 - 400) + 400, r, g, b);
}
/**
* 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
*/
static void wavelengthToRGB(double wavelength, unsigned char& r, unsigned char& g, unsigned char& b) {
double x, y, z;
cie1931WavelengthToXYZFit(wavelength, x, y, z);
double dr, dg, db;
srgbXYZ2RGB(x, y, z, dr, dg, db);
r = static_cast<unsigned char>(static_cast<int>(dr * 0xFF) & 0xFF);
g = static_cast<unsigned char>(static_cast<int>(dg * 0xFF) & 0xFF);
b = static_cast<unsigned char>(static_cast<int>(db * 0xFF) & 0xFF);
}
/**
* 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]
*/
static void srgbXYZ2RGB(double x, double y, double z, double& r, double& g, double& b) {
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;
r = srgbXYZ2RGBPostprocess(rl);
g = srgbXYZ2RGBPostprocess(gl);
b = srgbXYZ2RGBPostprocess(bl);
}
/**
* helper function for {@link #srgbXYZ2RGB(double[])}
*/
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 * std::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]
*/
static void cie1931WavelengthToXYZFit(double wavelength, double& x, double& y, double& z) {
double wave = wavelength;
{
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 * std::exp(-0.5 * t1 * t1)
+ 1.056 * std::exp(-0.5 * t2 * t2)
- 0.065 * std::exp(-0.5 * t3 * t3);
}
{
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 * std::exp(-0.5 * t1 * t1)
+ 0.286 * std::exp(-0.5 * t2 * t2);
}
{
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 * std::exp(-0.5 * t1 * t1)
+ 0.681 * std::exp(-0.5 * t2 * t2);
}
}
};
} //namespace
#endif
从 375nm 到 725nm 的颜色图如下所示：
这种方法的一个问题是它只能在 400-700nm 之间工作，在此范围之外它会急剧下降到黑色。另一个问题是较窄的蓝色。
作为比较，以下是来自 maxmax.com 的 Vision FAQ 的颜色：
我用它来可视化深度图，其中每个像素代表以米为单位的深度值，如下所示：
方法二
这是由 Aeash Partow 作为bitmap_image单个文件仅头文件库的一部分实现的：
inline rgb_t convert_wave_length_nm_to_rgb(const double wave_length_nm)
{
// Credits: Dan Bruton http://www.physics.sfasu.edu/astro/color.html
double red = 0.0;
double green = 0.0;
double blue = 0.0;
if ((380.0 <= wave_length_nm) && (wave_length_nm <= 439.0))
{
red = -(wave_length_nm - 440.0) / (440.0 - 380.0);
green = 0.0;
blue = 1.0;
}
else if ((440.0 <= wave_length_nm) && (wave_length_nm <= 489.0))
{
red = 0.0;
green = (wave_length_nm - 440.0) / (490.0 - 440.0);
blue = 1.0;
}
else if ((490.0 <= wave_length_nm) && (wave_length_nm <= 509.0))
{
red = 0.0;
green = 1.0;
blue = -(wave_length_nm - 510.0) / (510.0 - 490.0);
}
else if ((510.0 <= wave_length_nm) && (wave_length_nm <= 579.0))
{
red = (wave_length_nm - 510.0) / (580.0 - 510.0);
green = 1.0;
blue = 0.0;
}
else if ((580.0 <= wave_length_nm) && (wave_length_nm <= 644.0))
{
red = 1.0;
green = -(wave_length_nm - 645.0) / (645.0 - 580.0);
blue = 0.0;
}
else if ((645.0 <= wave_length_nm) && (wave_length_nm <= 780.0))
{
red = 1.0;
green = 0.0;
blue = 0.0;
}
double factor = 0.0;
if ((380.0 <= wave_length_nm) && (wave_length_nm <= 419.0))
factor = 0.3 + 0.7 * (wave_length_nm - 380.0) / (420.0 - 380.0);
else if ((420.0 <= wave_length_nm) && (wave_length_nm <= 700.0))
factor = 1.0;
else if ((701.0 <= wave_length_nm) && (wave_length_nm <= 780.0))
factor = 0.3 + 0.7 * (780.0 - wave_length_nm) / (780.0 - 700.0);
else
factor = 0.0;
rgb_t result;
const double gamma = 0.8;
const double intensity_max = 255.0;
#define round(d) std::floor(d + 0.5)
result.red = static_cast<unsigned char>((red == 0.0) ? red : round(intensity_max * std::pow(red * factor, gamma)));
result.green = static_cast<unsigned char>((green == 0.0) ? green : round(intensity_max * std::pow(green * factor, gamma)));
result.blue = static_cast<unsigned char>((blue == 0.0) ? blue : round(intensity_max * std::pow(blue * factor, gamma)));
#undef round
return result;
}
375-725nm 的波长图如下所示：
所以这在 400-725nm 更有用。当我可视化与方法 1 中相同的深度图时，我得到了以下信息。这些黑线有一个明显的问题，我认为这表明该代码中的小错误，我没有深入研究。在这种方法中，紫罗兰色有点窄，这会导致远处物体的对比度降低。
于 2018-08-23T08:49:31.100 回答
0
将波长的 CIExy 向 D65 白色投射到 sRGB 色域上
#!/usr/bin/ghci
ångstrømsfromTHz terahertz = 2997924.58 / terahertz
tristimulusXYZfromÅngstrøms å=map(sum.map(stimulus))[
[[1056,5998,379,310],[362,4420,160,267],[-65,5011,204,262]],
[[821,5688,469,405],[286,5309,163,311]],
[[1217,4370,118,360],[681,4590,260,138]]]
where stimulus[ω,μ,ς,σ]=ω/1000*exp(-((å-μ)/if å<μ then ς else σ)^2/2)
standardRGBfromTristimulusXYZ xyz=
map(gamma.sum.zipWith(*)(gamutConfine xyz))[
[3.2406,-1.5372,-0.4986],[-0.9689,1.8758,0.0415],[0.0557,-0.2040,1.057]]
gamma u=if u<=0.0031308 then 12.92*u else (u**(5/12)*211-11)/200
[red,green,blue,black]=
[[0.64,0.33],[0.3,0.6],[0.15,0.06],[0.3127,0.3290,0]]
ciexyYfromXYZ xyz=if xyz!!1==0 then black else map(/sum xyz)xyz
cieXYZfromxyY[x,y,l]=if y==0 then black else[x*l/y,l,(1-x-y)*l/y]
gamutConfine xyz=last$xyz:[cieXYZfromxyY[x0+t*(x1-x0),y0+t*(y1-y0),xyz!!1]|
x0:y0:_<-[black],x1:y1:_<-[ciexyYfromXYZ xyz],i<-[0..2],
[x2,y2]:[x3,y3]:_<-[drop i[red,green,blue,red]],
det<-[(x0-x1)*(y2-y3)-(y0-y1)*(x2-x3)],
t <-[((x0-x2)*(y2-y3)-(y0-y2)*(x2-x3))/det|det/=0],0<=t,t<=1]
sRGBfromÅ=standardRGBfromTristimulusXYZ.tristimulusXYZfromÅngstrøms
x s rgb=concat["\ESC[48;2;",
intercalate";"$map(show.(17*).round.(15*).max 0.min 1)rgb,
"m",s,"\ESC[49m"]
spectrum=concatMap(x" ".sRGBfromÅ)$takeWhile(<7000)$iterate(+60)4000
main=putStrLn spectrum
于 2020-01-08T23:54:35.473 回答