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

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
}

其他回答

对于快速和肮脏,你可以做到

import java.awt.Color;
private Color dropPrecision(Color c,int threshold){
    return new Color((c.getRed()/threshold),
                     (c.getGreen()/threshold),
                     (c.getBlue()/threshold));
}
public boolean inThreshold(Color _1,Color _2,int threshold){
    return dropPrecision(_1,threshold)==dropPrecision(_2,threshold);
}

利用整数除法对颜色进行量化。

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。

如果你有两个颜色对象c1和c2,你可以比较c1和c2的每个RGB值。

int diffRed   = Math.abs(c1.getRed()   - c2.getRed());
int diffGreen = Math.abs(c1.getGreen() - c2.getGreen());
int diffBlue  = Math.abs(c1.getBlue()  - c2.getBlue());

你可以将这些值除以饱和度的差异(255),你就会得到两者之间的差异。

float pctDiffRed   = (float)diffRed   / 255;
float pctDiffGreen = (float)diffGreen / 255;
float pctDiffBlue   = (float)diffBlue  / 255;

之后你就可以找到平均色差的百分比。

(pctDiffRed + pctDiffGreen + pctDiffBlue) / 3 * 100

这就得到了c和c之间的百分比差。

Android for ColorUtils API RGBToHSL 我有两个int argb颜色(color1, color2),我想要得到两种颜色之间的距离/差异。这是我所做的;

private float getHue(int color) {
    int R = (color >> 16) & 0xff;
    int G = (color >>  8) & 0xff;
    int B = (color      ) & 0xff;
    float[] colorHue = new float[3];
    ColorUtils.RGBToHSL(R, G, B, colorHue);
    return colorHue[0];
}

然后我使用下面的代码来查找两种颜色之间的距离。

private float getDistance(getHue(color1), getHue(color2)) {
    float avgHue = (hue1 + hue2)/2;
    return Math.abs(hue1 - avgHue);
}

Actually I walked the same path a couple of months ago. There is no perfect answer to the question (that was asked here a couple of times) but there is one, more sophisticated than the sqrt(r-r) etc. answer and more easy to implement directly with RGB without moving to all kinds of alternate color spaces. I found this formula here which is a low cost approximation of the quite complicated real formula (by the CIE which is the W3C of colors, since this is a not finished quest, you can find older and simpler color difference equations there). Good Luck.

编辑:为了子孙后代,这里是相关的C代码:

typedef struct {
     unsigned char r, g, b;
} RGB;

double ColourDistance(RGB e1, RGB e2)
{
    long rmean = ( (long)e1.r + (long)e2.r ) / 2;
    long r = (long)e1.r - (long)e2.r;
    long g = (long)e1.g - (long)e2.g;
    long b = (long)e1.b - (long)e2.b;
    return sqrt((((512+rmean)*r*r)>>8) + 4*g*g + (((767-rmean)*b*b)>>8));
}