又是每周一次的坑爹OPENCV！加油奥里给!

1、图像缩放——直接调用函数操作

    Mat img = imread("E:/lena.jpg");
    int img_cols = img.cols * 0.5;
    int img_rows = img.cols * 0.5;
    Mat change = Mat::zeros(img_cols,img_rows,CV_8UC3);
    resize(img, change, change.size());
    imshow("change",change);
    waitKey(0);
    return 0;

作用是缩放为原本图像的1/2，来看效果图

 

 姑且命名为大lena和小lena。

2、图像缩放——自己动手写，最近邻域插值法

原理：

目标上的每一个点都来自于原图像 

newX= x*(src 行/目标 ) newX =1*(10/5)=2 

newY= y*(src 列/目标 ) newY =1*(10/5)=2 

12.3 -> 12

 (以下内容转载自http://ddrv.cn/a/114636)

void nearestIntertoplation(cv::Mat& src, cv::Mat& dst, const int rows, const int cols)
{
    //比例尺
    const double scale_row = static_cast<double>(src.rows) / rows;
    const double scale_col = static_cast<double>(src.rows) / cols;

    //扩展src到dst
    dst = cv::Mat(rows, cols, src.type());
    assert(src.channels() == 1 && dst.channels() == 1);

    for (int i = 0; i < rows; ++i)//dst的行
        for (int j = 0; j < cols; ++j)//dst的列
        {
            //求插值的四个点
            double y = (i + 0.5) * scale_row + 0.5;
            double x = (j + 0.5) * scale_col + 0.5;
            int x1 = static_cast<int>(x);//col对应x
            if (x1 >= (src.cols - 2)) x1 = src.cols - 2;//防止越界
            int x2 = x1 + 1;
            int y1 = static_cast<int>(y);//row对应y
            if (y1 >= (src.rows - 2))  y1 = src.rows - 2;
            int y2 = y1 + 1;
            //根据目标图像的像素点（浮点坐标）找到原始图像中的4个像素点，取距离该像素点最近的一个原始像素值作为该点的值。
            assert(0 < x2 && x2 < src.cols && 0 < y2 && y2 < src.rows);
            std::vector<double> dist(4);
            dist[0] = distance(x, y, x1, y1);
            dist[1] = distance(x, y, x2, y1);
            dist[2] = distance(x, y, x1, y2);
            dist[3] = distance(x, y, x2, y2);

            int min_val = dist[0];
            int min_index = 0;
            for (int i = 1; i < dist.size(); ++i)
                if (min_val > dist[i])
                {
                    min_val = dist[i];
                    min_index = i;
                }

            switch (min_index)
            {
            case 0:
                dst.at<uchar>(i, j) = src.at<uchar>(y1, x1);
                break;
            case 1:
                dst.at<uchar>(i, j) = src.at<uchar>(y1, x2);
                break;
            case 2:
                dst.at<uchar>(i, j) = src.at<uchar>(y2, x1);
                break;
            case 3:
                dst.at<uchar>(i, j) = src.at<uchar>(y2, x2);
                break;
            default:
                assert(false);
            }
        }
}

double distance(const double x1, const double y1, const double x2, const double y2)//两点之间距离，这里用欧式距离
{
    return (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2);//只需比较大小，返回距离平方即可
}

实验结果如下

 

3、图像裁切

    cv::Mat img = cv::imread("E:/lena.jpg", 0);
    if (img.empty()) return -1;
    int col = img.cols;
    int row = img.rows;
    Rect rect(col * 1 / 4, col*1/2, row * 1 / 4, row*1/2);
    Mat image = img(rect);
    imshow("sb", image);
    waitKey(0);
    return 0;

最终结果

 

 4、图像的位移

    cv::Mat img = cv::imread("E:/lena.jpg", 0);
    if (img.empty()) return -1;
    int col = img.cols;
    int row = img.rows;
    Mat dst;
    dst.create(row, col, img.type());
    Vec3b* p;
    for (int i = 0; i < row; i++)
    {
        p = dst.ptr<Vec3b>(i);
        for (int j = 0; j < col; j++)
        {
            int x = j - 10;
            int y = i - 10;
            if (x >= 0 && y >= 0 && x < col && y < row)
            {
                img.at<cv::Vec3b>(i, j) = dst.ptr<cv::Vec3b>(y)[x];
            }
        }
    }
    waitKey(0);
    return 0;