c++ - OpenCV 的 SIFT 匹配数作为第一个关键点数

Question

使用此代码，当我打印匹配项数时，我得到的数字与第一张图像中找到的关键点数相同...例如一个输出可以是

关键点数量1:682

关键点数量2:159

匹配数(FLANN):682

匹配数(蛮力):682

第二张图片中的匹配点数怎么可能比关键点数多？我希望匹配的数量小于或等于两个图像的关键点数量之间的最小值。

cv::SiftFeatureDetector  detector;
    detector = SiftFeatureDetector(
                                    0, // nFeatures
                                    4, // nOctaveLayers
                                    0.04, // contrastThreshold
                                    10, //edgeThreshold
                                    1.6 //sigma
                                    );


    std::vector<KeyPoint> keypoints1, keypoints2;

    int dxP = gray1.rows / 10;

    cv::Mat mask = cv::Mat::zeros(gray1.size(), CV_8U); // all 0
    mask(Rect(dxP, dxP, gray1.cols - 2*dxP, gray1.rows - 2*dxP)) = 1;

    detector.detect(gray1, keypoints1, mask);
    cout << "number of keypoints1: " << keypoints1.size() << endl;

    detector.detect(gray2, keypoints2, mask);
    cout << "number of keypoints2: " << keypoints2.size() << endl;

    SiftDescriptorExtractor extractor;
    Mat descriptors1, descriptors2;
    extractor.compute(gray1, keypoints1, descriptors1);
    extractor.compute(gray2, keypoints2, descriptors2);

    //Matching descriptor vectors using FLANN matcher
    FlannBasedMatcher matcher;
    std::vector< DMatch > matches;
    matcher.match(descriptors1, descriptors2, matches);
    cout << "number of matches (FLANN): " << matches.size() << endl;

    // Matching descriptor vectors using Brute Force matcher
    BFMatcher BFmatcher(NORM_L2);
    vector<DMatch> BFmatches;
    BFmatcher.match(descriptors1, descriptors2, BFmatches);

    cout << "number of matches (Brute Force): " << BFmatches.size() << endl;

Answer 1

这取决于参数的顺序，以实现您的需要反转顺序:

matcher.match(descriptors1, descriptors2, matches);

到

matcher.match(descriptors2, descriptors1, matches);

注意使用的顺序，甚至 OpenCV 文档也混淆了 matcher.match 调用的顺序，后来颠倒了 matches[i].trainIdx 和 匹配[i].queryIdx

::match 的第一个参数指的是 queryDescriptors(即 queryIdx)