7

是否有连接两个（或更多）附近轮廓的功能？看看我的输入/输出，你就会明白我的意思了……</p>

我的代码：

[... some processing ...]

// getting contours
std::vector<std::vector<cv::Point> > contours;
findContours(input, contours, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_SIMPLE);

// approximate contours
std::vector<std::vector<cv::Point> > contours_poly( contours.size() );
for( int i = 0; i < contours.size(); i++ ) {
  approxPolyDP(cv::Mat(contours[i]), contours_poly[i], 5, true );
}

// debugging
cv::Scalar colors[3];
colors[0] = cv::Scalar(255, 0, 0);
colors[1] = cv::Scalar(0, 255, 0);
colors[2] = cv::Scalar(0, 0, 255);
for (int idx = 0; idx < contours_poly.size(); idx++) {
  cv::drawContours(output, contours_poly, idx, colors[idx % 3]);
}

输出 输出

4

2 回答 2

11

我想出了这个解决方案，因为我只需要整个对象周围的边界框：

[... some processing ...]

// getting contours
std::vector<std::vector<cv::Point> > contours;
findContours(input, contours, CV_RETR_EXTERNAL, CV_CHAIN_APPROX_SIMPLE);

// approximate contours
std::vector<std::vector<cv::Point> > contours_poly( contours.size() );
for( int i = 0; i < contours.size(); i++ ) {
  approxPolyDP(cv::Mat(contours[i]), contours_poly[i], 5, true );
}

// merge all contours into one vector
std::vector<cv::Point> merged_contour_points;
for (int i = 0; i < contours_poly.size(); i++) {
  for (int j = 0; j < contours_poly[i].size(); j++) {
    merged_contour_points.push_back(contours_poly[i][j]);
  }
}

// get rotated bounding box
std::vector<cv::Point> hull;
cv::convexHull(cv::Mat(merged_contour_points),hull);
cv::Mat hull_points(hull);
cv::RotatedRect rotated_bounding_rect = minAreaRect(hull_points);

有时去除胡椒噪音可以带来更好的结果：

void removePepperNoise(cv::Mat &mask)
{
    for ( int y=2; y<mask.rows-2; y++ ) {
        uchar *pUp2 = mask.ptr(y-2);
        uchar *pUp1 = mask.ptr(y-1);
        uchar *pThis = mask.ptr(y);
        uchar *pDown1 = mask.ptr(y+1);
        uchar *pDown2 = mask.ptr(y+2);
        pThis += 2;
        pUp1 += 2;
        pUp2 += 2;
        pDown1 += 2;
        pDown2 += 2;

        for (int x=2; x<mask.cols-2; x++) {
            uchar value = *pThis; // Get this pixel value (0 or 255). // Check if this is a black pixel that is surrounded by white pixels
            if (value == 0) {
                bool above, left, below, right, surroundings;
                above = *(pUp2 - 2) && *(pUp2 - 1) && *(pUp2) && *(pUp2 + 1) && *(pUp2 + 2);
                left = *(pUp1 - 2) && *(pThis - 2) && *(pDown1 - 2);
                below = *(pDown2 - 2) && *(pDown2 - 1) && *(pDown2) && *(pDown2 + 1) && *(pDown2 + 2);
                right = *(pUp1 + 2) && *(pThis + 2) && *(pDown1 + 2);
                surroundings = above && left && below && right;
                if (surroundings == true) {
                    // Fill the whole 5x5 block as white. Since we know
                    // the 5x5 borders are already white, we just need to
                    // fill the 3x3 inner region.
                    *(pUp1 - 1) = 255;
                    *(pUp1 + 0) = 255;
                    *(pUp1 + 1) = 255;
                    *(pThis - 1) = 255;
                    *(pThis + 0) = 255;
                    *(pThis + 1) = 255;
                    *(pDown1 - 1) = 255;
                    *(pDown1 + 0) = 255;
                    *(pDown1 + 1) = 255;
                    // Since we just covered the whole 5x5 block with
                    // white, we know the next 2 pixels won't be black,
                    // so skip the next 2 pixels on the right.
                    pThis += 2;
                    pUp1 += 2;
                    pUp2 += 2;
                    pDown1 += 2;
                    pDown2 += 2;
                }
            }
            // Move to the next pixel on the right.
            pThis++;
            pUp1++;
            pUp2++;
            pDown1++;
            pDown2++;
        }
    }
}
于 2012-01-25T11:09:27.043 回答
2

只需通过点并找到最近的起点或终点，然后将它们连接起来。在您的情况下，很难决定是否应该连接轮廓。如果像 Adrian Popovici 所说的形态学没有帮助，您必须指定一些最大距离来决定是否要连接点。

于 2012-01-25T07:22:47.613 回答