非连接形态滤波器

经过一些简单的预处理后,我收到了分段图像的布尔掩码.

我想"增强"面具的边框,使它们更光滑.为此我使用具有相当大的圆内核的OPEN形态滤波器,它非常有效,直到分段对象之间的距离足够.但在很多样本中,物体粘在一起.是否存在一些或多或少的简单方法来平滑这种图像而不改变其形态？

如果不首先应用形态滤镜,您可以尝试检测图像的外部轮廓.现在,您可以将这些外部轮廓绘制为填充轮廓,然后应用形态滤镜.这是有效的,因为现在你没有任何填充孔.这很简单.

另一种方法:

找到外部轮廓

取轮廓点的坐标x,y.您可以将它们视为1-D信号,并对这些信号应用平滑滤波器

在下面的代码中,我将第二种方法应用于样本图像.

输入图像

外部轮廓没有任何平滑

在对x和y 1-D信号应用高斯滤波器之后

C++代码

Mat im = imread("4.png", 0);

Mat cont = im.clone();
Mat original = Mat::zeros(im.rows, im.cols, CV_8UC3);
Mat smoothed = Mat::zeros(im.rows, im.cols, CV_8UC3);

// contour smoothing parameters for gaussian filter
int filterRadius = 5;
int filterSize = 2 * filterRadius + 1;
double sigma = 10;      

vector > contours;
vector hierarchy;
// find external contours and store all contour points
findContours(cont, contours, hierarchy, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_NONE, Point(0, 0));
for(size_t j = 0; j < contours.size(); j++)
{
    // draw the initial contour shape
    drawContours(original, contours, j, Scalar(0, 255, 0), 1);
    // extract x and y coordinates of points. we'll consider these as 1-D signals
    // add circular padding to 1-D signals
    size_t len = contours[j].size() + 2 * filterRadius;
    size_t idx = (contours[j].size() - filterRadius);
    vector x, y;
    for (size_t i = 0; i < len; i++)
    {
        x.push_back(contours[j][(idx + i) % contours[j].size()].x);
        y.push_back(contours[j][(idx + i) % contours[j].size()].y);
    }
    // filter 1-D signals
    vector xFilt, yFilt;
    GaussianBlur(x, xFilt, Size(filterSize, filterSize), sigma, sigma);
    GaussianBlur(y, yFilt, Size(filterSize, filterSize), sigma, sigma);
    // build smoothed contour
    vector > smoothContours;
    vector smooth;
    for (size_t i = filterRadius; i < contours[j].size() + filterRadius; i++)
    {
        smooth.push_back(Point(xFilt[i], yFilt[i]));
    }
    smoothContours.push_back(smooth);

    drawContours(smoothed, smoothContours, 0, Scalar(255, 0, 0), 1);

    cout << "debug contour " << j << " : " << contours[j].size() << ", " << smooth.size() << endl;
}