如何创建Display P3颜色的不饱和版本？

Question

我有一个颜色UIColor(displayP3Red: 1, green: 0, blue: 0.8, alpha: 1)。我想创建一个不饱和的版本 - 具有相同的色调和亮度，但较少的饱和度，如原始饱和度的一半。我该怎么做呢？

Answer 1

extension UIColor {  
    // Calling this with 0.5 as argument returns a color whose saturation is 50% of that of the receiver. 
    func desaturatedBy(fraction: CGFloat) -> UIColor { 
    var hue: CGFloat = 0 
    var saturation: CGFloat = 0 
    var brightness: CGFloat = 0 
    var alpha: CGFloat = 0 
    let success = getHue(&hue, saturation: &saturation, brightness: &brightness, alpha: &alpha) 
    assert(success) 

    saturation *= fraction 

    return UIColor(hue: hue, saturation: saturation, brightness: brightness, alpha: alpha) 
    } 
} 

这工作，因为第e getter在扩展的sRGB色彩空间中返回HSB，而UIColor的初始化程序也在扩展的sRGB色彩空间中获取HSB。所以我们不需要做任何色彩空间转换。

Answer 2

考虑到您可以计算DCI-P3的归一化主矩阵（NPM），它应该很简单：NPM的中间一行代表亮度因子。我将举例说明这一点使用Colour，它假设你正在使用线性值：

import numpy as np 
import colour 

# Computing the sRGB Luminance Equation, you should be familiar with the 
# resulting Luminance factors. 
print(colour.RGB_luminance_equation(
    colour.sRGB_COLOURSPACE.primaries, 
    colour.sRGB_COLOURSPACE.whitepoint)) 
# Y = 0.212639005872(R) + 0.715168678768(G) + 0.0721923153607(B) 

# Computing the DCI-P3 Luminance Equation. 
print(colour.RGB_luminance_equation(
    colour.DCI_P3_COLOURSPACE.primaries, 
    colour.DCI_P3_COLOURSPACE.whitepoint)) 
# Y = 0.209491677913(R) + 0.721595254161(G) + 0.0689130679262(B) 

# Computing Luminance of given RGB colour, this is assuming it is representing linear values. 
DCI_P3_LUMINANCE_FACTORS = np.array([0.209491677913, 0.721595254161, 0.0689130679262]) 
RGB = np.array([1.0, 0.0, 0.8]) 
Y = np.dot(RGB, DCI_P3_LUMINANCE_FACTORS) 
print(Y) 
# 0.264622132254 

随着Ÿ代表您给出颜色的亮度，如果你想在50％饱和度，你可以做这样的事情：

线性插值（RGB，[Y，Y，Y]，0.5）