計算兩種 LaB 顏色之間的距離 Java

Question

我正在開發一個程序，我試圖在其中找到兩種 LaB 顏色之間的歐幾里德距離。 我想知道他們計算兩者之間距離的方式是否正確。 這是我在做什么：

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);  
}

Answer 1

因此，對於任何想知道兩個 Lab 值之間的歐幾里得距離之間是否存在細微差異的人來說，並沒有。 正如維基所說

... Lab* 中任意兩種顏色之間的相對感知差異可以通過將每種顏色視為三維空間中的一個點（具有三個分量：L、a、b*）並取它們之間的歐幾里德距離來近似

這個問題的解決方案是這樣的。

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));    
}

Answer 2

為了在Lab色彩空間中獲得更准確的感知距離，您可以使用 CIEDE2000 (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;
}
}