填充后存储图像的问题

Posted 2023-02-21

【中文标题】填充后存储图像的问题

【英文标题】：problem with storing an image after padding

【发布时间】：2020-08-12 22:00:41

【问题描述】：

我正在使用 5x5 窗口执行简单的屏蔽操作。按照这篇文章Problem with Operation on Border pixels of an image（第二个解决方案）中接受的解决方案，我创建了一个图像img_clamp，它可以用作具有4个额外行和4个额外列的缓冲区。请在下面找到示例代码。

int main(int argc, char** argv)

    Mat input = imread("C:/Users/20181217/Desktop/images/imgs/den_check.png");
    //input.rows = 256, input.cols =512
    Mat output = (input.rows,input.cols,input.type()); //row and col = 256 and 512 
    
        
    //number of additional rows and columns you woluld ike on each side
    int top, left, right, bottom;
    top = 2;
    left = 2;
    right = 2;
    bottom = 2;
    
    //define new image with additional borders
    Mat img_clamp(input.rows + 4, input.cols + 4, CV_8UC3);  
        
    
    //if you want to replicate the border of the image
    copyMakeBorder(input, img_clamp, top, left, right, bottom, BORDER_REPLICATE);
    //img_clamp row and col size : 260 and 516


//Now you can access the image without having to worry about the borders as shown below

//start iterationg from the 2nd row till 258th row (this leaves 1,2,259 and 260th rows for the out of bounds access by the 5x5 window) 
for(int i=2;i<input.rows-2;i++)
 
    //start iteration from the 2nd col till 514th col (this leaves 1,2,515 and 516th cols for the out of bounds access by the 5x5 window)
  for(int j=2;i<input.cols-2;i++)
   
     
    temp_red = img_clamp.at<Vec3b>[0](i-2,j-2) + img_clamp.at<Vec3b>[0](i-2,j+2) + img_clamp.at<Vec3b>[0](i+2,j-2) + img_clamp.at<Vec3b>[0](i+2,j+2);
    temp_green = img_clamp.at<Vec3b>[1](i-2,j-2) + img_clamp.at<Vec3b>[1](i-2,j+2) + img_clamp.at<Vec3b>[1](i+2,j-2) + img_clamp.at<Vec3b>[1](i+2,j+2);
    temp_blue = img_clamp.at<Vec3b>[2](i-2,j-2) + img_clamp.at<Vec3b>[2](i-2,j+2) + img_clamp.at<Vec3b>[2](i+2,j-2) + img_clamp.at<Vec3b>[2](i+2,j+2);
     
     ...
     
     //store values in the output image which has the same the size as input image (i.e 256 and 512 (rows and cols))
     output.at<Vec3b>[0](i-2,j-2) =temp_red
     output.at<Vec3b>[1](i-2,j-2) =temp_green
     output.at<Vec3b>[2](i-2,j-2) =temp_blue
         
         
    
  

//Code for checking the output is matching with the golden data(ideal image)<---- the ideal image is from a different language halide, My goal is to replicate the same logic in c++

Mat diff = abs(ideal-output);

    Mat diff = abs(small_img - ideal);
    //cout << diff;
    int r, g, b, t, n,r_b,g_b,b_b;
    r = 0; //faulty red
    g = 0;//faulty green
    b = 0;//faulty blue
    n = 393216; //total number of pixels
    //_b  indicates the border pixels
    r_b = 0;
    b_b = 0;
    g_b = 0;

    t = 1;//threshold for the difference

          // printing out the difference between final image and ideal image (if any)
    for (int i = 0; i < input.cols; i++)//512 col size x
    
        for (int j = 0; j < input.rows; j++)//256 row size y
        
            if ((int)diff.at<Vec3b>(j, i)[0] > t)
            
                if (i == 508 || i == 509 || i == 510 || i == 511 || j == 252 || j == 253 || j == 254 || j == 255) //border pixels(right most and bottom most of the image)
                    r_b++;//increment if its one of the border pixels
                //printing the pixel position which has the wrong value
                cout << "problem at (" << j << "," << i << ")  of red :" << (int)output.at<Vec3b>(j, i)[0] << ", expected value:" << (int)ideal.at<Vec3b>(j, i)[0] << endl;
                r++;//increment if the red pixel is faulty and not matching with the ideal image
                n--;
            
            if ((int)diff.at<Vec3b>(j, i)[1] > t)
            
                if (i == 508 || i == 509 || i == 510 || i == 511 || j == 252 || j == 253 || j == 254 || j == 255)
                    g_b++;
                
                g++;//increment if the green pixel is faulty and not matching with the ideal image
                n--;

            
            if ((int)diff.at<Vec3b>(j, i)[2] > t)
            
                if (i == 508 || i == 509 || i == 510 || i == 511 || j == 252 || j == 253 || j == 254 || j == 255)
                    b_b++;              

                b++;//increment if the blue pixel is faulty and not matching with the ideal image
                n--;
            

        
    
    cout << endl << endl << "for a threshold of difference greater than " << t << endl << "total faulty pixels are(R,G,B) :(" << r << " " << g << " " << b << ")"<<endl;
    cout << " The border pixels present in the faulty pixels are :(" << r_b << " " << g_b << " " << b_b << endl << endl;
    cout<<" correct pixels that match with the ideal image = " << n;
    
    return 0;

我想将输出存储为常规图像**（未填充）**。 除了最后四行和四列之外，一切都按照逻辑进行。控制台的输出是

我知道在存储时(i-2,j-2) 出错了，但我无法找到解决方法。

任何关于如何解决它的帮助/建议都将受到高度赞赏。

提前致谢

【问题讨论】：

您添加了 4 行/列，但在创建 img_clamp 时只添加了 2。

@1201ProgramAlarm，上两下两下共4行

但是在img_clamp的构造中，你只是将2添加到输入Mat的行数和列数中。

@1201ProgramAlarm，感谢您指出这一点，我从 3x3 掩码过滤器中提取代码并对其进行了编辑，所以我错过了那部分。我已经更新了

@CrisLuengo 我已经更新了信息，当我试图详细解释时，我发现哪里出了问题。这是 for 循环中的一个小错误，我使用“input.rows”而不是“img_clamp.rows”导致最后 4 行和列的问题，现在已解决。感谢您的建议:-)

【参考方案1】：

问题在于for loop 的边界 而不是这个

for(int i=2;i<input.rows-2;i++)//size =256
 
    //start iteration from the 2nd col till 514th col (this leaves 1,2,515 and 516th cols for the out of bounds access by the 5x5 window)
  for(int j=2;i<input.cols-2;i++)//size =512

应该是的

for(int i=2;i<img_clamp.rows-2;i++)//size = 260
 
    //start iteration from the 2nd col till 514th col (this leaves 1,2,515 and 516th cols for the out of bounds access by the 5x5 window)
  for(int j=2;i<img_clamp.cols-2;i++)//size = 516