java - 使RGB获得最接近的色彩算法更准确

我已经创建了一个包含名称和rgb值的巨大颜色列表（花了很长时间），现在我创建了一个算法，可以将相应的颜色获取到最接近的值。
它似乎工作得很好，但有时当有一个奇怪的值，它得到了错误的颜色。
示例输出

Log: InputRGB: R:7.1009636 | G:83.84344 | B:2.5013387
Log: ColorToCompare: Ball Blue (R13.0,G67.0,B80.0) CLOSE:0.4588677 | CurrentColor: Acid Green CLOSE: 0.41585693
Log: ColorToCompare: Bitter Lemon (R79.0,G88.0,B5.0) CLOSE:0.5143066 | CurrentColor: Ball Blue CLOSE: 0.4588677
Log: ColorToCompare: Citrine (R89.0,G82.0,B4.0) CLOSE:0.5610447 | CurrentColor: Bitter Lemon CLOSE: 0.5143066
Log: ColorToCompare: Smoky Black (R6.0,G5.0,B3.0) CLOSE:0.57945675 | CurrentColor: Citrine CLOSE: 0.5610447
Log: ColorName:Smoky Black
Log: End Color: R:6.0 G:5.0 B:3.0
Log: InputRGB:    R:7.1009636 | G:83.84344 | B:2.5013387

我创建的用于计算此值的代码：

   public String getClosetColor(float red, float green, float blue){

        Functions.log("InputRGB: R:" + red + " | G:" + green + " | B:" + blue);

        Color lastColor = null;
        for(Color eachColor : this.colors)
        {
            if(lastColor == null){
                lastColor = eachColor;
            }

            float lastColorCloseness = (getClose(red, lastColor.red) + getClose(green, lastColor.green) + getClose(blue, lastColor.blue)) / 3f;
            float thisColorCloseness = (getClose(red, eachColor.red) + getClose(green, eachColor.green) + getClose(blue, eachColor.blue)) / 3f;

            if(Float.isFinite(thisColorCloseness) && Float.isFinite(lastColorCloseness))
            {
                //If they are the same, choose a random one.
                if(lastColorCloseness == thisColorCloseness){
                    if(MathUtils.random() > 0.5f){
                        lastColor = eachColor;
                    }
                }
                //If this one is greater then set it.
                else if(thisColorCloseness > lastColorCloseness){
                    Functions.log(
                            "ColorToCompare: " + eachColor.nameOfColor + " (R" + eachColor.red + ",G" + eachColor.green + ",B" + eachColor.blue + ") CLOSE:" + thisColorCloseness +
                                    " | CurrentColor: " + lastColor.nameOfColor + " CLOSE: " + lastColorCloseness
                    );

                    lastColor = eachColor;
                }
            }
        }

        Functions.log("ColorName:" + lastColor.nameOfColor);
        Functions.log("End Color: R:" + lastColor.red + " G:" + lastColor.green + " B:" + lastColor.blue);
        Functions.log("InputRGB:    R:" + red + " | G:" + green + " | B:" + blue);

        return "";
    }

    //Basically if one is higher than the other then devide by it.
    private float getClose(float firstNumber, float secondNumber){
        if(firstNumber < secondNumber){
            return firstNumber / secondNumber;
        }
        else{
            return secondNumber / firstNumber;
        }
    }

最佳答案

我不知道你是怎么想到距离函数的，但有点尴尬让我解释一下：
使用颜色的比率而不是差异，例如：

float lastColorCloseness = (getClose(red, lastColor.red) + getClose(green, lastColor.green) + getClose(blue, lastColor.blue)) / 3f;

这有一种奇怪的效果，即对距离相等的颜色不应用相同的颜色。例如比较
col1(100, 50, 200)使用col2(50, 100, 150)和col3(150, 100, 250)。
假设col2和col3与col1之间的距离等于：

abs(100-50)+abs(50-100)+abs(200-150)=150
abs(100-150)+abs(50-100)+abs(200-250)=150

你的距离函数给出了不同的结果：

(50/100+50/100+150/250)/3=0.53
(50/100+50/100+200/250)/3=0.6

正如“大卫华莱士”所说，这并不是最夸张的结果。
改用欧几里德函数之类的距离函数。

关于java - 使RGB获得最接近的色彩算法更准确，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/25620837/