怀旧滤镜实现原理
不管是荣耀华为手机还是其他的手机,我们都可以找到相机中的怀旧效果,这是手机中常用的一种滤镜效果。
怀旧风格的设计主要是在图像的颜色空间进行处理。以BGR为例,对B、G、R这3个通道的颜色数值进行处理,让图像有一种泛黄的怀旧效果。设计的转换公式如下:
B=0.272r+0.534g+0.131*b
G=0.349r+0.686g+0.168*b
R=0.393r+0.769g+0.189*b
计算公式中的小写的bgr是原图像的RGB通道的颜色,结果BGR是怀旧变换后的值。需要注意的是,颜色值的范围在[0,255],需要在程序中约束一下。
实现怀旧滤镜
既然我们已经了解了其实现的原理公式。下面我们直接上代码实现该功能,具体代码如下所示:
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def cowboy_effect(img): new_img = img.copy() h, w, n = img.shape for i in range (w): for j in range (h): b = img[j, i, 0 ] g = img[j, i, 1 ] r = img[j, i, 2 ] B = int ( 0.272 * r + 0.534 * g + 0.131 * b) G = int ( 0.349 * r + 0.686 * g + 0.168 * b) R = int ( 0.393 * r + 0.769 * g + 0.189 * b) new_img[j, i, 0 ] = max ( 0 , min (B, 255 )) new_img[j, i, 1 ] = max ( 0 , min (G, 255 )) new_img[j, i, 2 ] = max ( 0 , min (R, 255 )) return new_img if __name__ = = "__main__" : img = cv2.imread( "48.jpg" ) cv2.imshow( "0" , img) cv2.imshow( "1" , cowboy_effect(img)) cv2.waitKey() cv2.destroyAllWindows() |
运行之后,效果如下:
连环画滤镜原理
从怀旧滤镜就可以看出来,其实相机的各种滤镜效果就是对RGB的颜色通道进行计算处理。既然怀旧滤镜有公式,那么肯定的连环画滤镜也有公式。它的具体公式如下:
R = |g – b + g + r| * r / 256
G = |b – g + b + r| * r / 256
B = |b – g + b + r| * g / 256
实现连环画滤镜
有了公式,下面直接套用公式即可。具体代码如下所示:
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# 连环画滤镜 def comics_effect(img): new_img = img.copy() h, w, n = img.shape for i in range (w): for j in range (h): b = img[j, i, 0 ] g = img[j, i, 1 ] r = img[j, i, 2 ] R = int ( int ( abs (g - b + g + r)) * r / 256 ) G = int ( int ( abs (b - g + b + r)) * r / 256 ) B = int ( int ( abs (b - g + b + r)) * g / 256 ) new_img[j, i, 0 ] = R new_img[j, i, 1 ] = G new_img[j, i, 2 ] = B return new_img if __name__ = = "__main__" : img = cv2.imread( "48.jpg" ) cv2.imshow( "0" , img) cv2.imshow( "1" , comics_effect(img)) cv2.waitKey() cv2.destroyAllWindows() |
运行之后,效果如下:
综上所述,基本上所有的基础滤镜都是通过对RGB通道的颜色值进行公式计算得到的。当然,要是数学很好,又对算法情有独钟的读者,可以自己自研滤镜算法丰富滤镜的效果。
熔铸算法
r = r*128/(g+b +1);
g = g*128/(r+b +1);
b = b*128/(g+r +1);
冰冻算法
r = (r-g-b)*3/2;
g = (g-r-b)*3/2;
b = (b-g-r)*3/2;
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#include <math.h> #include <opencv/cv.h> #include <opencv/highgui.h> #define MAXSIZE (32768) using namespace cv; using namespace std; void casting(const Mat& src) { Mat img; src.copyTo(img); int width = src.cols; int heigh = src.rows; Mat dst(img.size(),CV_8UC3); for ( int y = 0 ;y<heigh;y + + ) { uchar * imgP = img.ptr<uchar>(y); uchar * dstP = dst.ptr<uchar>(y); for ( int x = 0 ;x<width;x + + ) { float b0 = imgP[ 3 * x]; float g0 = imgP[ 3 * x + 1 ]; float r0 = imgP[ 3 * x + 2 ]; float b = b0 * 255 / (g0 + r0 + 1 ); float g = g0 * 255 / (b0 + r0 + 1 ); float r = r0 * 255 / (g0 + b0 + 1 ); r = (r> 255 ? 255 : (r< 0 ? 0 : r)); g = (g> 255 ? 255 : (g< 0 ? 0 : g)); b = (b> 255 ? 255 : (b< 0 ? 0 : b)); dstP[ 3 * x] = (uchar)b; dstP[ 3 * x + 1 ] = (uchar)g; dstP[ 3 * x + 2 ] = (uchar)r; } } imshow( "熔铸" ,dst); imwrite( "D:/img/熔铸.jpg" ,dst); } void freezing(const Mat& src) { Mat img; src.copyTo(img); int width = src.cols; int heigh = src.rows; Mat dst(img.size(),CV_8UC3); for ( int y = 0 ;y<heigh;y + + ) { uchar * imgP = img.ptr<uchar>(y); uchar * dstP = dst.ptr<uchar>(y); for ( int x = 0 ;x<width;x + + ) { float b0 = imgP[ 3 * x]; float g0 = imgP[ 3 * x + 1 ]; float r0 = imgP[ 3 * x + 2 ]; float b = (b0 - g0 - r0) * 3 / 2 ; float g = (g0 - b0 - r0) * 3 / 2 ; float r = (r0 - g0 - b0) * 3 / 2 ; r = (r> 255 ? 255 : (r< 0 ? - r : r)); g = (g> 255 ? 255 : (g< 0 ? - g : g)); b = (b> 255 ? 255 : (b< 0 ? - b : b)); / / r = (r> 255 ? 255 : (r< 0 ? 0 : r)); / / g = (g> 255 ? 255 : (g< 0 ? 0 : g)); / / b = (b> 255 ? 255 : (b< 0 ? 0 : b)); dstP[ 3 * x] = (uchar)b; dstP[ 3 * x + 1 ] = (uchar)g; dstP[ 3 * x + 2 ] = (uchar)r; } } imwrite( "D:/img/冰冻.jpg" ,dst); } int main() { Mat src = imread( "D:/img/scene04.jpg" , 1 ); imshow( "src" ,src); casting(src); freezing(src); waitKey(); } |
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原文链接:https://liyuanjinglyj.blog.csdn.net/article/details/115118465