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。
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
}
我在我的android up中使用了这个,它似乎令人满意,尽管不建议使用RGB空间:
public double colourDistance(int red1,int green1, int blue1, int red2, int green2, int blue2)
{
double rmean = ( red1 + red2 )/2;
int r = red1 - red2;
int g = green1 - green2;
int b = blue1 - blue2;
double weightR = 2 + rmean/256;
double weightG = 4.0;
double weightB = 2 + (255-rmean)/256;
return Math.sqrt(weightR*r*r + weightG*g*g + weightB*b*b);
}
然后我用下面的方法得到相似度的百分比:
double maxColDist = 764.8339663572415;
double d1 = colourDistance(red1,green1,blue1,red2,green2,blue2);
String s1 = (int) Math.round(((maxColDist-d1)/maxColDist)*100) + "% match";
它工作得很好。
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。
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));
}