Question

我正在使用Sobel屏蔽边缘检测而不使用任何特殊库。我想要的输出是512x512矩阵的文本文件，其值介于0到1之间。我已经通过设置较小的值（如50而不是＆＃39; ROW-2＆＃39;）来检查代码是否正常工作。和＆＃39; COL-2＆＃39;。但是，如果我把它们放回去，代码将永远运行。

常量值为：

const int ROW = 512;
const int COL = 512;
const double Gx [3][3] = { {-1.0,0.0,1.0},{-2.0,0.0,2.0},{-1.0,0.0,1.0}};
const double Gy [3][3] = { {1.0,2.0,1.0},{0.0,0.0,0.0},{-1.0,-2.0,-1.0}};

这是主要功能：

int main()

{  
    double NewImage[ROW][COL] = {0};    

    for (int i = 0; i < ROW; i++)
    {
        for (int j = 0; j < COL; j++)
        {
            NewImage[i][j] = 0;
        }
    }

    for (int i = 0; i < ROW-2; i++)
    {
        for (int j = 0; j < COL-2; j++)
        {

            NewImage[i+1][j+1] = SobelConvolution(i,j); 
        }
    }

    ofstream newImage;
    string filename;
    filename = "output image.txt";

    newImage.open (filename.c_str());

    for(int rows = 0; rows < ROW; rows++)
    {
        for(int cols = 0; cols < COL; cols++)
        {
            newImage << NewImage[ROW][COL] <<" ";
        }
        newImage << endl;
    }

    newImage.close();

    return 0;
}

这是SobelConvolution的功能：

double SobelConvolution(int row, int col)
{   
    double convX;
    double convY;
    double conv;

    convX = ImageReader(row,col)*Gx[2][2]
            + ImageReader(row,col+1)*Gx[2][1]
            + ImageReader(row,col+2)*Gx[2][0]
            + ImageReader(row+1,col)*Gx[1][2]
            + ImageReader(row+1,col+1)*Gx[1][1]
            + ImageReader(row+1,col+2)*Gx[1][0]
            + ImageReader(row+2,col)*Gx[0][2]
            + ImageReader(row+2,col+1)*Gx[0][1]
            + ImageReader(row+2,col+2)*Gx[0][0];

    convY = ImageReader(row,col)*Gy[2][2]
            + ImageReader(row,col+1)*Gy[2][1]
            + ImageReader(row,col+2)*Gy[2][0]
            + ImageReader(row+1,col)*Gy[1][2]
            + ImageReader(row+1,col+1)*Gy[1][1]
            + ImageReader(row+1,col+2)*Gy[1][0]
            + ImageReader(row+2,col)*Gy[0][2]
            + ImageReader(row+2,col+1)*Gy[0][1]
            + ImageReader(row+2,col+2)*Gy[0][0];

    conv = sqrt((convX*convX) + (convY*convY));


    return conv;
}

这是函数ImageReader：

double ImageReader(int r, int c)
{
    double OrigImage[ROW][COL];

    ifstream defaultImage ("image.txt");

    if (defaultImage.good())
    {
        for (int i = 0; i < ROW; i++)
        {
            for (int j = 0; j < COL; j++)
            {
                defaultImage >> OrigImage[i][j];
            }
        }
    }
    return OrigImage [r][c]; 
}

任何提示或建议？提前谢谢！

Answer 1

以下是一些注释：

<强> ImageReader

只返回数组的一个值，不需要在每个时间读取整个数组，只需要一个值。在我看来，这个功能是多余的。

<强> SobelConvolution

此功能很好，但有一个不必要的变量 - conv。

<强> main

当我们已经NewImage时，我不知道为什么要将0的每个值初始化为0！你也不知道实际上需要NewImage

这是我要写的内容（广泛评论）：

double SobelConvolution(int row, int col) { //ImageReader has been removed, it was unnecessary. The code has been moved here double oldImage[ROW][COL]; std::ifstream defaultImage{ "image.txt" }; //Error handling if file doesn't exist - consider doing something else :) if (!defaultImage.is_open()) return 0; //Initialize array for (int i = 0; i < ROW; ++i) for (int j = 0; j < COL; ++j) defaultImage >> oldImage[i][j]; //You should always declare variables where they are first used, this //reduces the possibility of errors //We can just access the array directly double convX = oldImage[row][col] * Gx[2][2] + oldImage[row][col + 1] * Gx[2][1] + oldImage[row][col + 2] * Gx[2][0] + oldImage[row + 1][col] * Gx[1][2] + oldImage[row + 1][col + 1] * Gx[1][1] + oldImage[row + 1][col + 2] * Gx[1][0] + oldImage[row + 2][col] * Gx[0][2] + oldImage[row + 2][col + 1] * Gx[0][1] + oldImage[row + 2][col + 2] * Gx[0][0]; double convY = oldImage[row][col] * Gy[2][2] + oldImage[row][col + 1] * Gy[2][1] + oldImage[row][col + 2] * Gy[2][0] + oldImage[row + 1][col] * Gy[1][2] + oldImage[row + 1][col + 1] * Gy[1][1] + oldImage[row + 1][col + 2] * Gy[1][0] + oldImage[row + 2][col] * Gy[0][2] + oldImage[row + 2][col + 1] *Gy[0][1] + oldImage[row + 2][col + 2]*Gy[0][0]; //No need to create a separate variable just to return it return sqrt((convX*convX) + (convY*convY)); } int main() { //= {} Initializes every element to 0, you don't need to do it :) Just so you know :) //Note that it crashes here, because my stack size was too small, //maybe consider using a dynamic array (512 * 512 is pretty big) :) //double NewImage[ROW][COL] = {}; //The array is not really needed, see below std::string filename = "oimage.txt"; std::ofstream newImage{ filename }; //No need to create another array just to output it again, //Just output the calculated values - this doesn't ignore the first/last values for (int rows = 0; rows < ROW; rows++) { for (int cols = 0; cols < COL; cols++) newImage << SobelConvolution(rows, cols) << " "; newImage << '\n'; //std::endl flushes the stream, while \n does not - it is faster :) } newImage.close(); return 0; }

Answer 2

你真的想打开单个图像文件18次并读取每行和每列的所有数据只是为了返回单行和列18次？为什么不一次读取图像文件并将图像数据数组传递给函数？

Answer 3

你所做的不仅仅是效率低下，而且是完全疯狂的。

对于图像的每个像素，您调用SobelConvolution，后者又调用ImageReader 18次（其中6次没有用，因为相应的系数为零）。但可怕的是，ImageReader每次都会从文本文件中执行完整的图像读取，只需简单的数组查找即可。

总而言之，您正在执行4718592文件流打开/关闭以及从文件读取1236950581248值，其中只需1次打开/关闭和262144次读取就足够了。（不计算单个读取比直接数组访问要贵得多。）完整运行可能持续两个小时或更长时间。

我不明白为什么我的c ++代码运行得这么慢

3 个答案: