我想设计一个程序，可以帮助我在5种预定义的颜色中评估哪一种更类似于可变颜色，以及与可变颜色的百分比。问题是我不知道如何手动一步一步地做到这一点。所以想一个程序就更难了。
更多细节:颜色来自不同颜色的管子和凝胶的照片。我有5个不同颜色的管子，每个代表5个等级中的1个。我想拍摄其他样本的照片，然后在电脑上通过比较颜色来评估样本属于哪个级别，我也想知道一个近似的百分比。我想要一个这样做的程序:http://www.colortools.net/color_matcher.html
如果你能告诉我该采取什么步骤，即使它们需要我手动思考和执行。那会很有帮助的。
当前回答
请参阅维基百科关于色差的文章以获得正确的线索。 基本上，你想要在多维颜色空间中计算一个距离度量。
但是RGB并不是“感知上一致的”，所以Vadim建议的欧几里得RGB距离度量将与人类感知的颜色之间的距离不匹配。首先，L*a*b*是一个感知上均匀的颜色空间，delta度量是常用的。但有更精致的色彩空间和更精致的delta公式，更接近人类的感知。
你需要学习更多关于颜色空间和光源的知识来进行转换。但如果想要一个比欧几里得RGB度量更好的快速公式，只需这样做:
假设你的RGB值在sRGB颜色空间中 找到sRGB到L*a*b*的转换公式 将sRGB颜色转换为L*a*b* 计算两个L*a*b*值之间的delta
计算成本不高，只是一些非线性公式和一些乘法和加法。
其他回答
我猜你最后想分析一幅完整的图像，对吧?所以你可以检查单位颜色矩阵的最小/最大差值。
大多数处理图形的数学操作都使用矩阵，因为使用矩阵的可能算法通常比经典的逐点距离和比较计算更快。(例如，对于使用DirectX, OpenGL，…的操作)
所以我认为你应该从这里开始:
http://en.wikipedia.org/wiki/Identity_matrix
http://en.wikipedia.org/wiki/Matrix_difference_equation
…正如Beska在上面评论的那样:
这可能不会带来最好的“可见”差异……
这也意味着，如果你在处理图像，你的算法取决于你对“相似”的定义。
Kotlin版本与你想匹配的百分比有多少。
方法调用，参数为percent
isMatchingColor(intColor1, intColor2, 95) // should match color if 95% similar
方法体
private fun isMatchingColor(intColor1: Int, intColor2: Int, percent: Int = 90): Boolean {
val threadSold = 255 - (255 / 100f * percent)
val diffAlpha = abs(Color.alpha(intColor1) - Color.alpha(intColor2))
val diffRed = abs(Color.red(intColor1) - Color.red(intColor2))
val diffGreen = abs(Color.green(intColor1) - Color.green(intColor2))
val diffBlue = abs(Color.blue(intColor1) - Color.blue(intColor2))
if (diffAlpha > threadSold) {
return false
}
if (diffRed > threadSold) {
return false
}
if (diffGreen > threadSold) {
return false
}
if (diffBlue > threadSold) {
return false
}
return true
}
只是另一个答案，尽管它与Supr的答案相似-只是不同的颜色空间。
问题是:人类感知颜色的差异并不均匀，而RGB颜色空间忽略了这一点。因此，如果你使用RGB颜色空间，只是计算两种颜色之间的欧几里得距离，你可能会得到一个在数学上绝对正确的差异，但与人类告诉你的不一致。
This may not be a problem - the difference is not that large I think, but if you want to solve this "better" you should convert your RGB colors into a color space that was specifically designed to avoid the above problem. There are several ones, improvements from earlier models (since this is based on human perception we need to measure the "correct" values based on experimental data). There's the Lab colorspace which I think would be the best although a bit complicated to convert it to. Simpler would be the CIE XYZ one.
这里有一个网站列出了在不同颜色空间之间转换的公式，所以你可以尝试一下。
以下所有方法的结果都是0-100。
internal static class ColorDifference
{
internal enum Method
{
Binary, // true or false, 0 is false
Square,
Dimensional,
CIE76
}
public static double Calculate(Method method, int argb1, int argb2)
{
int[] c1 = ColorConversion.ArgbToArray(argb1);
int[] c2 = ColorConversion.ArgbToArray(argb2);
return Calculate(method, c1[1], c2[1], c1[2], c2[2], c1[3], c2[3], c1[0], c2[0]);
}
public static double Calculate(Method method, int r1, int r2, int g1, int g2, int b1, int b2, int a1 = -1, int a2 = -1)
{
switch (method)
{
case Method.Binary:
return (r1 == r2 && g1 == g2 && b1 == b2 && a1 == a2) ? 0 : 100;
case Method.CIE76:
return CalculateCIE76(r1, r2, g1, g2, b1, b2);
case Method.Dimensional:
if (a1 == -1 || a2 == -1) return Calculate3D(r1, r2, g1, g2, b1, b2);
else return Calculate4D(r1, r2, g1, g2, b1, b2, a1, a2);
case Method.Square:
return CalculateSquare(r1, r2, g1, g2, b1, b2, a1, a2);
default:
throw new InvalidOperationException();
}
}
public static double Calculate(Method method, Color c1, Color c2, bool alpha)
{
switch (method)
{
case Method.Binary:
return (c1.R == c2.R && c1.G == c2.G && c1.B == c2.B && (!alpha || c1.A == c2.A)) ? 0 : 100;
case Method.CIE76:
if (alpha) throw new InvalidOperationException();
return CalculateCIE76(c1, c2);
case Method.Dimensional:
if (alpha) return Calculate4D(c1, c2);
else return Calculate3D(c1, c2);
case Method.Square:
if (alpha) return CalculateSquareAlpha(c1, c2);
else return CalculateSquare(c1, c2);
default:
throw new InvalidOperationException();
}
}
// A simple idea, based on on a Square
public static double CalculateSquare(int argb1, int argb2)
{
int[] c1 = ColorConversion.ArgbToArray(argb1);
int[] c2 = ColorConversion.ArgbToArray(argb2);
return CalculateSquare(c1[1], c2[1], c1[2], c2[2], c1[3], c2[3]);
}
public static double CalculateSquare(Color c1, Color c2)
{
return CalculateSquare(c1.R, c2.R, c1.G, c2.G, c1.B, c2.B);
}
public static double CalculateSquareAlpha(int argb1, int argb2)
{
int[] c1 = ColorConversion.ArgbToArray(argb1);
int[] c2 = ColorConversion.ArgbToArray(argb2);
return CalculateSquare(c1[1], c2[1], c1[2], c2[2], c1[3], c2[3], c1[0], c2[0]);
}
public static double CalculateSquareAlpha(Color c1, Color c2)
{
return CalculateSquare(c1.R, c2.R, c1.G, c2.G, c1.B, c2.B, c1.A, c2.A);
}
public static double CalculateSquare(int r1, int r2, int g1, int g2, int b1, int b2, int a1 = -1, int a2 = -1)
{
if (a1 == -1 || a2 == -1) return (Math.Abs(r1 - r2) + Math.Abs(g1 - g2) + Math.Abs(b1 - b2)) / 7.65;
else return (Math.Abs(r1 - r2) + Math.Abs(g1 - g2) + Math.Abs(b1 - b2) + Math.Abs(a1 - a2)) / 10.2;
}
// from:http://stackoverflow.com/questions/9018016/how-to-compare-two-colors
public static double Calculate3D(int argb1, int argb2)
{
int[] c1 = ColorConversion.ArgbToArray(argb1);
int[] c2 = ColorConversion.ArgbToArray(argb2);
return Calculate3D(c1[1], c2[1], c1[2], c2[2], c1[3], c2[3]);
}
public static double Calculate3D(Color c1, Color c2)
{
return Calculate3D(c1.R, c2.R, c1.G, c2.G, c1.B, c2.B);
}
public static double Calculate3D(int r1, int r2, int g1, int g2, int b1, int b2)
{
return Math.Sqrt(Math.Pow(Math.Abs(r1 - r2), 2) + Math.Pow(Math.Abs(g1 - g2), 2) + Math.Pow(Math.Abs(b1 - b2), 2)) / 4.41672955930063709849498817084;
}
// Same as above, but made 4D to include alpha channel
public static double Calculate4D(int argb1, int argb2)
{
int[] c1 = ColorConversion.ArgbToArray(argb1);
int[] c2 = ColorConversion.ArgbToArray(argb2);
return Calculate4D(c1[1], c2[1], c1[2], c2[2], c1[3], c2[3], c1[0], c2[0]);
}
public static double Calculate4D(Color c1, Color c2)
{
return Calculate4D(c1.R, c2.R, c1.G, c2.G, c1.B, c2.B, c1.A, c2.A);
}
public static double Calculate4D(int r1, int r2, int g1, int g2, int b1, int b2, int a1, int a2)
{
return Math.Sqrt(Math.Pow(Math.Abs(r1 - r2), 2) + Math.Pow(Math.Abs(g1 - g2), 2) + Math.Pow(Math.Abs(b1 - b2), 2) + Math.Pow(Math.Abs(a1 - a2), 2)) / 5.1;
}
/**
* Computes the difference between two RGB colors by converting them to the L*a*b scale and
* comparing them using the CIE76 algorithm { http://en.wikipedia.org/wiki/Color_difference#CIE76}
*/
public static double CalculateCIE76(int argb1, int argb2)
{
return CalculateCIE76(Color.FromArgb(argb1), Color.FromArgb(argb2));
}
public static double CalculateCIE76(Color c1, Color c2)
{
return CalculateCIE76(c1.R, c2.R, c1.G, c2.G, c1.B, c2.B);
}
public static double CalculateCIE76(int r1, int r2, int g1, int g2, int b1, int b2)
{
int[] lab1 = ColorConversion.ColorToLab(r1, g1, b1);
int[] lab2 = ColorConversion.ColorToLab(r2, g2, b2);
return Math.Sqrt(Math.Pow(lab2[0] - lab1[0], 2) + Math.Pow(lab2[1] - lab1[1], 2) + Math.Pow(lab2[2] - lab1[2], 2)) / 2.55;
}
}
internal static class ColorConversion
{
public static int[] ArgbToArray(int argb)
{
return new int[] { (argb >> 24), (argb >> 16) & 0xFF, (argb >> 8) & 0xFF, argb & 0xFF };
}
public static int[] ColorToLab(int R, int G, int B)
{
// http://www.brucelindbloom.com
double r, g, b, X, Y, Z, fx, fy, fz, xr, yr, zr;
double Ls, fas, fbs;
double eps = 216.0f / 24389.0f;
double k = 24389.0f / 27.0f;
double Xr = 0.964221f; // reference white D50
double Yr = 1.0f;
double Zr = 0.825211f;
// RGB to XYZ
r = R / 255.0f; //R 0..1
g = G / 255.0f; //G 0..1
b = B / 255.0f; //B 0..1
// assuming sRGB (D65)
if (r <= 0.04045) r = r / 12;
else r = (float)Math.Pow((r + 0.055) / 1.055, 2.4);
if (g <= 0.04045) g = g / 12;
else g = (float)Math.Pow((g + 0.055) / 1.055, 2.4);
if (b <= 0.04045) b = b / 12;
else b = (float)Math.Pow((b + 0.055) / 1.055, 2.4);
X = 0.436052025f * r + 0.385081593f * g + 0.143087414f * b;
Y = 0.222491598f * r + 0.71688606f * g + 0.060621486f * b;
Z = 0.013929122f * r + 0.097097002f * g + 0.71418547f * b;
// XYZ to Lab
xr = X / Xr;
yr = Y / Yr;
zr = Z / Zr;
if (xr > eps) fx = (float)Math.Pow(xr, 1 / 3.0);
else fx = (float)((k * xr + 16.0) / 116.0);
if (yr > eps) fy = (float)Math.Pow(yr, 1 / 3.0);
else fy = (float)((k * yr + 16.0) / 116.0);
if (zr > eps) fz = (float)Math.Pow(zr, 1 / 3.0);
else fz = (float)((k * zr + 16.0) / 116);
Ls = (116 * fy) - 16;
fas = 500 * (fx - fy);
fbs = 200 * (fy - fz);
int[] lab = new int[3];
lab[0] = (int)(2.55 * Ls + 0.5);
lab[1] = (int)(fas + 0.5);
lab[2] = (int)(fbs + 0.5);
return lab;
}
}
快速回答
我找到这个帖子是因为我需要这个问题的Swift版本。由于还没有人给出答案，我的答案是:
extension UIColor {
var rgba: (red: CGFloat, green: CGFloat, blue: CGFloat, alpha: CGFloat) {
var red: CGFloat = 0
var green: CGFloat = 0
var blue: CGFloat = 0
var alpha: CGFloat = 0
getRed(&red, green: &green, blue: &blue, alpha: &alpha)
return (red, green, blue, alpha)
}
func isSimilar(to colorB: UIColor) -> Bool {
let rgbA = self.rgba
let rgbB = colorB.rgba
let diffRed = abs(CGFloat(rgbA.red) - CGFloat(rgbB.red))
let diffGreen = abs(rgbA.green - rgbB.green)
let diffBlue = abs(rgbA.blue - rgbB.blue)
let pctRed = diffRed
let pctGreen = diffGreen
let pctBlue = diffBlue
let pct = (pctRed + pctGreen + pctBlue) / 3 * 100
return pct < 10 ? true : false
}
}
用法:
let black: UIColor = UIColor.black
let white: UIColor = UIColor.white
let similar: Bool = black.isSimilar(to: white)
我设置小于10%的差异返回相似的颜色，但你可以自定义这自己。
