我正在尝试找到最适合加载图像的颜色并将其应用于背景中。适应图像,让UI感觉更自然。 到目前为止我找到了2个方案:
1>平均像素(下面的代码):
final Color acclimatizeAverage(BufferedImage img) {
long avgr = 0, avgb = 0, avgg = 0;
for (int i = 0; i < img.getWidth(); i++) {
for (int j = 0; j < img.getHeight(); j++) {
Color c = new Color(img.getRGB(i, j));
avgr += c.getRed(); avgb += c.getBlue(); avgg += c.getGreen();
}
}
avgr = (avgr/(img.getHeight()*img.getWidth()));
avgg = (avgg/(img.getHeight()*img.getWidth()));
avgb = (avgb/(img.getHeight()*img.getWidth()));
Color c = new Color((int)avgr, (int)avgg, (int)avgb);
return c;
}
2 - ;将像素分组为颜色的固定区间(代码如下):
Map<Color, Integer> createBins() {
Map<Color, Integer> bins = new HashMap<>();
bins.put(Color.red, 0);
bins.put(Color.magenta, 0);
bins.put(Color.orange, 0);
bins.put(Color.PINK, 0);
bins.put(Color.yellow, 0);
bins.put(Color.LIGHT_GRAY, 0);
bins.put(Color.GREEN, 0);
bins.put(Color.GRAY, 0);
bins.put(Color.DARK_GRAY, 0);
bins.put(Color.CYAN, 0);
bins.put(Color.BLUE, 0);
bins.put(Color.BLACK, 0);
return bins;
}
int compare(Color a, Color b) {
return (int)Math.sqrt((a.getRed() - b.getRed())*(a.getRed() - b.getRed())
+ (a.getBlue() - b.getBlue())*(a.getBlue() - b.getBlue())
+ (a.getGreen()- b.getGreen())*(a.getGreen()- b.getGreen()));
}
BufferedImage acclimatizeGrouping(BufferedImage img) {
Map<Color, Integer> bins = createBins();
for (int i = 0; i < img.getWidth(); i++) {
int min = Integer.MAX_VALUE; Color minC = null;
for (int j = 0; j < img.getHeight(); j++) {
Color c = new Color(img.getRGB(i, j));
for (Map.Entry<Color, Integer> entry : bins.entrySet()) {
Integer integer = compare(entry.getKey(), c);
if(integer < min) {
min = integer;
minC = entry.getKey();
}
}
bins.put(minC, bins.get(minC)+1);
}
}
int max = -1, n = 1; Color c = null;
for (Map.Entry<Color, Integer> entry : bins.entrySet()) {
Integer integer = entry.getValue();
if(integer > max) {
max = integer;
c = entry.getKey();
}
}
return c;
}
但分组正在产生奇怪的结果....
左侧是分组结果产生的颜色,右侧是图像
为什么会产生这样的结果???
averaing正在产生更正确的结果:
答案 0 :(得分:1)
我认为问题在于RGB不是人类欧几里德空间。您使用欧氏距离来比较颜色,但它不利于人类的颜色感。有关详细信息,请参阅this link。
编辑:更准确地说,你应该使用这个算法:
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));
}
答案 1 :(得分:0)
此问题是,您的compare(Color a, Color b)
方法未正确实现,可以使用Math.pow()
方法进行一些基本的重构。
以编程方式查找相似颜色的基本公式是
((r2-r1) 2 +(g2-g1) 2 +(b2-b1) 2 ) 1 / 2
应用于Java,导致修改后的compare(Color a, Color b)
int compare(Color a, Color b){
return Math.sqrt(( Math.pow( b.getRed() - a.getRed() )
+ ( Math.pow( b.getGreen() - a.getGreen() )
+ ( Math.pow( b.getBlue() - a.getBlue() ));
}