Question

我试图用下面的代码用双线性插值对正弦/余弦进行失真，我保持与旧图像相同的新图像大小。结果看起来并不好。

Mat biliniar(Mat old, int freq){
namedWindow( "car2", CV_WINDOW_AUTOSIZE );
imshow( "car2", old); 
int height = old.size().height;
int width = old.size().width;

unsigned char R1,R2,R3,R4;     // Colours at the four neighbours
unsigned char G1,G2,G3,G4;
unsigned char B1,B2,B3,B4;
double RT1, GT1, BT1;          // Interpolated colours at T1 and T2
double RT2, GT2, BT2;
unsigned char R,G,B;           // Final colour at a destination pixel
unsigned char *dst;            // Destination image - must be allocated here! 
int x,y;               // Coordinates on destination image
double fi,fj;              // Corresponding coordinates on source image

Mat res(height,width,CV_8UC3,cv::Scalar(100));
for (int i = 0; i < width; i++) {
    for (int j = 0; j < height; j++) {
        fi = i;
        fj =  (j - (((height/30)*(sin((M_PI*i)/(freq/2)))))) + sqrt(height) ; 
        //cout << fj << " ";
        if(fj > 0 &  fj < height){
            //cout << " p1: x:" << floor(fi) << " y:" << floor(fj);
            //cout << " p2: x:" << ceil(fi) << " y:" << floor(fj);
            //cout << " p3: x:" << floor(fi) << " y:" << ceil(fj);
            //cout << " p4: x:" << ceil(fi) << " y:" << ceil(fj);
            Vec3b color1 = old.at<Vec3b>(Point(floor(fi)-1,floor(fj)));
            R1 = color1.val[0]; G1 = color1.val[1]; B1 = color1.val[2];
            Vec3b color2 = old.at<Vec3b>(Point(ceil(fi)+1,floor(fj)));
            R2 = color2.val[0]; G2 = color2.val[1]; B2 = color2.val[2];
            Vec3b color3 = old.at<Vec3b>(Point(floor(fi)-1,ceil(fj)));
            R3 = color3.val[0]; G3 = color3.val[1]; B3 = color3.val[2];
            Vec3b color4 = old.at<Vec3b>(Point(ceil(fi)+1,ceil(fj)));
            R4 = color4.val[0]; G4 = color4.val[1]; B4 = color4.val[2];
            RT1 = (R1+R2)/2; GT1 = (G1+G2)/2; BT1 = (B1+B2)/2;
            RT2 = (R3+R4)/2; GT2 = (G3+G4)/2; BT2 = (B3+B4)/2;
            R=(unsigned char)(RT1+RT2);
            G=(unsigned char)(GT1+GT2);
            B=(unsigned char)(BT1+BT2);
            Vec3b finalColor(R,G,B);
            res.at<Vec3b>(j,i) = finalColor;
        }
    }
}
namedWindow( "bilinear", CV_WINDOW_AUTOSIZE );
imshow( "bilinear", res); 
return res;

}

最近邻插值的结果更好，任何人都知道为什么？

由于

Answer 1

你实际上并没有在该代码中进行任何双线性插值，只是将所有四个像素加在一起。

您需要计算fi和fj的小数部分并使用它们来执行插值，例如：

double fi_part = fi - floor(fi);
double fj_part = fj - floor(fj);

// perform interpolation in i direction:
RT1 = (R1*(1.0-fi_part)) + (R2*fi_part);
GT1 = (G1*(1.0-fi_part)) + (G2*fi_part);
BT1 = (B1*(1.0-fi_part)) + (B2*fi_part);
RT2 = (R3*(1.0-fi_part)) + (R4*fi_part);
GT2 = (G3*(1.0-fi_part)) + (G4*fi_part);
BT2 = (B3*(1.0-fi_part)) + (B4*fi_part);

// perform interpolation in j direction:
R=(unsigned char)(RT1*(1.0-fj_part)+RT2*fj_part);
G=(unsigned char)(GT1*(1.0-fj_part)+GT2*fj_part);
B=(unsigned char)(BT1*(1.0-fj_part)+BT2*fj_part);

Vec3b finalColor(R,G,B);
res.at<Vec3b>(j,i) = finalColor;

因此，当你移动一个纹素时，当fi_part从0.0增加到1.0时，R1，G1，B1越来越少，R2，G2，B2越来越多，将对RT1，GT1，BT1做出贡献。对于RT2等也是如此，对于涉及fj_part的计算也类似。

代码可以提高很多速度，但希望你能得到基本的想法。

用双线性插值c ++正弦图像失真

1 个答案: