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计算两种实验室颜色之间的距离Java

euclidean-distance colors java

ProgrammingCuber · 技术社区 · 7 年前

public static void main(String[] args) {

    double[] lab = {58.974604845047, 15.037506818771362, -64.0113115310669}; //blue
    double[] lab1 = {58.701420307159424, 14.014512300491333, -64.46481943130493};//blue

    double distance = euclideanDistance(lab, lab1);
    System.out.println("Lab: " + distance); 

}

private static double euclideanDistance(double[] lab , double []lab1){
    //L = 0 - 100 range
    //A = -86.185 - 98.254 range
    //B = 107.863 - 94.482 range
    double Lmean = (lab[0] + lab1[0]) / 2;
    double L = lab[0] - lab1[0];
    double A = lab[1] - lab1[1];
    double B = lab[2] - lab1[2];
    double weightL = 2 + Lmean /100;
    double weightA = 4.0;
    double weightB = 2 + (107.863 - Lmean) / 100;

    return Math.sqrt(weightL * L * L + weightA * A * A + weightB * B * B);  
}

2 回复 | 直到 7 年前

ProgrammingCuber 7 年前

所以,对于任何想知道两个实验室值之间的欧几里得距离是否有微小差异的人来说,这是不存在的。如所述 wiki

这个问题的解决方案是这样的。

private  double euclideanDistance(double[] lab , double []lab1){
    double L = lab[0] - lab1[0];
    double A = lab[1] - lab1[1];
    double B = lab[2] - lab1[2];
    return Math.sqrt((L * L) +  (A * A) +  (B * B));    
}

Vlad Badalyan 4 年前

为了获得更精确的感知距离 Lab 可以使用CIEDE2000的颜色空间(CIE Delta E 2000)

public class CIEDE2000 {

/**
 * Calculate the colour difference value between two colours in lab space.
 * @param L1 first colour's L component
 * @param a1 first colour's a component
 * @param b1 first colour's b component
 * @param L2 second colour's L component
 * @param a2 second colour's a component
 * @param b2 second colour's b component
 * @return the CIE 2000 colour difference
 */
public static double calculateDeltaE(double L1, double a1, double b1, double L2, double a2, double b2) {
    double Lmean = (L1 + L2) / 2.0;
    double C1 =  Math.sqrt(a1*a1 + b1*b1);
    double C2 =  Math.sqrt(a2*a2 + b2*b2);
    double Cmean = (C1 + C2) / 2.0;

    double G =  ( 1 - Math.sqrt( Math.pow(Cmean, 7) / (Math.pow(Cmean, 7) + Math.pow(25, 7)) ) ) / 2; //ok
    double a1prime = a1 * (1 + G);
    double a2prime = a2 * (1 + G);

    double C1prime =  Math.sqrt(a1prime*a1prime + b1*b1);
    double C2prime =  Math.sqrt(a2prime*a2prime + b2*b2);
    double Cmeanprime = (C1prime + C2prime) / 2;

    double h1prime =  Math.atan2(b1, a1prime) + 2*Math.PI * (Math.atan2(b1, a1prime)<0 ? 1 : 0);
    double h2prime =  Math.atan2(b2, a2prime) + 2*Math.PI * (Math.atan2(b2, a2prime)<0 ? 1 : 0);
    double Hmeanprime =  ((Math.abs(h1prime - h2prime) > Math.PI) ? (h1prime + h2prime + 2*Math.PI) / 2 : (h1prime + h2prime) / 2);

    double T =  1.0 - 0.17 * Math.cos(Hmeanprime - Math.PI/6.0) + 0.24 * Math.cos(2*Hmeanprime) + 0.32 * Math.cos(3*Hmeanprime + Math.PI/30) - 0.2 * Math.cos(4*Hmeanprime - 21*Math.PI/60);

    double deltahprime =  ((Math.abs(h1prime - h2prime) <= Math.PI) ? h2prime - h1prime : (h2prime <= h1prime) ? h2prime - h1prime + 2*Math.PI : h2prime - h1prime - 2*Math.PI);

    double deltaLprime = L2 - L1;
    double deltaCprime = C2prime - C1prime;
    double deltaHprime =  2.0 * Math.sqrt(C1prime*C2prime) * Math.sin(deltahprime / 2.0);
    double SL =  1.0 + ( (0.015*(Lmean - 50)*(Lmean - 50)) / (Math.sqrt( 20 + (Lmean - 50)*(Lmean - 50) )) );
    double SC =  1.0 + 0.045 * Cmeanprime;
    double SH =  1.0 + 0.015 * Cmeanprime * T;

    double deltaTheta =  (30 * Math.PI / 180) * Math.exp(-((180/Math.PI*Hmeanprime-275)/25)*((180/Math.PI*Hmeanprime-275)/25));
    double RC =  (2 * Math.sqrt(Math.pow(Cmeanprime, 7) / (Math.pow(Cmeanprime, 7) + Math.pow(25, 7))));
    double RT =  (-RC * Math.sin(2 * deltaTheta));

    double KL = 1;
    double KC = 1;
    double KH = 1;

    double deltaE = Math.sqrt(
            ((deltaLprime/(KL*SL)) * (deltaLprime/(KL*SL))) +
            ((deltaCprime/(KC*SC)) * (deltaCprime/(KC*SC))) +
            ((deltaHprime/(KH*SH)) * (deltaHprime/(KH*SH))) +
            (RT * (deltaCprime/(KC*SC)) * (deltaHprime/(KH*SH)))
            );

    return deltaE;
}
}