c++ - 使用 OpenCV 描述符匹配 findFundamentalMat

Question

我早些时候发布了关于同一程序的问题，但没有收到任何答复。我已经纠正了我当时遇到的问题，结果却遇到了一个新问题。

基本上，我使用未校准的方法自动校正旋转和平移的立体图像对。我使用 SURF 等特征检测算法在两幅图像(左右立体图像对)中找到点，然后再次使用 SURF 匹配两幅图像之间的点。然后我需要使用这些匹配点来找到我可以用来校正图像的基本矩阵。

我的问题是这样的。我的匹配点存储在描述符匹配的单个 vector 中，然后过滤异常值。 findFundamentalMat 将两个独立的匹配点数组作为输入。我不知道如何将 vector 转换为两个单独的数组。

cout << "< Matching descriptors..." << endl;
vector<DMatch> filteredMatches;
crossCheckMatching( descriptorMatcher, descriptors1, descriptors2, filteredMatches, 1 );
cout << filteredMatches.size() << " matches" << endl << ">" << endl;

The vector is created.

void crossCheckMatching( Ptr<DescriptorMatcher>& descriptorMatcher,
                         const Mat& descriptors1, const Mat& descriptors2,
                         vector<DMatch>& filteredMatches12, int knn=1 )
{
    filteredMatches12.clear();
    vector<vector<DMatch> > matches12, matches21;
    descriptorMatcher->knnMatch( descriptors1, descriptors2, matches12, knn );
    descriptorMatcher->knnMatch( descriptors2, descriptors1, matches21, knn );
    for( size_t m = 0; m < matches12.size(); m++ )
    {
        bool findCrossCheck = false;
        for( size_t fk = 0; fk < matches12[m].size(); fk++ )
        {
            DMatch forward = matches12[m][fk];

            for( size_t bk = 0; bk < matches21[forward.trainIdx].size(); bk++ )
            {
                DMatch backward = matches21[forward.trainIdx][bk];
                if( backward.trainIdx == forward.queryIdx )
                {
                    filteredMatches12.push_back(forward);
                    findCrossCheck = true;
                    break;
                }
            }
            if( findCrossCheck ) break;
        }
    }
}

The matches are cross checked and stored within filteredMatches.

cout << "< Computing homography (RANSAC)..." << endl;
vector<Point2f> points1; KeyPoint::convert(keypoints1, points1, queryIdxs);
vector<Point2f> points2; KeyPoint::convert(keypoints2, points2, trainIdxs);
H12 = findHomography( Mat(points1), Mat(points2), CV_RANSAC, ransacReprojThreshold );
cout << ">" << endl;

The homography is found based on a threshold which is set at run time in the command prompt.

//Mat drawImg;
if( !H12.empty() ) // filter outliers
{
    vector<char> matchesMask( filteredMatches.size(), 0 );
    vector<Point2f> points1; KeyPoint::convert(keypoints1, points1, queryIdxs);
    vector<Point2f> points2; KeyPoint::convert(keypoints2, points2, trainIdxs);
    Mat points1t; perspectiveTransform(Mat(points1), points1t, H12);
    for( size_t i1 = 0; i1 < points1.size(); i1++ )
    {
        if( norm(points2[i1] - points1t.at<Point2f>((int)i1,0)) < 4 ) // inlier
            matchesMask[i1] = 1;
    }
    /* draw inliers
    drawMatches( leftImg, keypoints1, rightImg, keypoints2, filteredMatches, drawImg, CV_RGB(0, 255, 0), CV_RGB(0, 0, 255), matchesMask, 2 ); */
}

The matches are further filtered to remove outliers.

...然后呢？我如何将剩下的东西分成两个 Mat 的匹配点以在 findFundamentalMat 中使用？

编辑

我现在已经使用我的掩码制作了一个 finalMatches vector (这取代了上面的最终过滤过程):

Mat drawImg;
if( !H12.empty() ) // filter outliers
{
    size_t i1;
    vector<char> matchesMask( filteredMatches.size(), 0 );
    vector<Point2f> points1; KeyPoint::convert(keypoints1, points1, queryIdxs);
    vector<Point2f> points2; KeyPoint::convert(keypoints2, points2, trainIdxs);
    Mat points1t; perspectiveTransform(Mat(points1), points1t, H12);
    for( i1 = 0; i1 < points1.size(); i1++ )
    {
        if( norm(points2[i1] - points1t.at<Point2f>((int)i1,0)) < 4 ) // inlier
            matchesMask[i1] = 1;
    }
    for( i1 = 0; i1 < filteredMatches.size(); i1++ )
    {
        if ( matchesMask[i1] == 1 )
            finalMatches.push_back(filteredMatches[i1]);
    }
    namedWindow("matches", 1);
    // draw inliers
    drawMatches( leftImg, keypoints1, rightImg, keypoints2, filteredMatches, drawImg, CV_RGB(0, 255, 0), CV_RGB(0, 0, 255), matchesMask, 2 );
    imshow("matches", drawImg);
}

但是我仍然不知道如何将我的 finalMatches DMatch vector 拆分为我需要输入 findFundamentalMat 的 Mat 数组，请帮忙!!!

编辑

工作(某种)解决方案:

Mat drawImg;
vector<Point2f> finalPoints1;
vector<Point2f> finalPoints2;
if( !H12.empty() ) // filter outliers
{
    size_t i, idx;
    vector<char> matchesMask( filteredMatches.size(), 0 );
    vector<Point2f> points1; KeyPoint::convert(keypoints1, points1, queryIdxs);
    vector<Point2f> points2; KeyPoint::convert(keypoints2, points2, trainIdxs);
    Mat points1t; perspectiveTransform(Mat(points1), points1t, H12);

    for( i = 0; i < points1.size(); i++ )
    {
        if( norm(points2[i] - points1t.at<Point2f>((int)i,0)) < 4 ) // inlier
            matchesMask[i] = 1;
    }

    for ( idx = 0; idx < filteredMatches.size(); idx++)
    {
        if ( matchesMask[idx] == 1 ) {
            finalPoints1.push_back(keypoints1[filteredMatches[idx].queryIdx].pt);
            finalPoints2.push_back(keypoints2[filteredMatches[idx].trainIdx].pt);
        }
    }    

    namedWindow("matches", 0);
    // draw inliers
    drawMatches( leftImg, keypoints1, rightImg, keypoints2, filteredMatches, drawImg, CV_RGB(0, 255, 0), CV_RGB(0, 0, 255), matchesMask, 2 );
    imshow("matches", drawImg);
}

然后我将 finalPoints1 和 finalPoints2 作为 Mat 的输入到 findFundamentalMat 中。现在我唯一的问题是我的输出没有像预期的那样远程，图像都搞砸了:-/

Answer 1

您的匹配数组是描述符数组的偏移量。由于每个描述符都有一个对应的关键点，您可以简单地从索引迭代并构建两个关键点数组。然后可以将这些关键点输入到 findFundamentalMat 中。

编辑:

我认为您的错误是在丢失信息的地方生成 finalMatches。 vector filteredMatches 过载。 matchesMask 为 1 的索引显示 keypoints1 的索引，而 finalMatches 中存储的索引是 keypoints2 的索引。通过缩小到 finalMatches，您实际上失去了第一组索引。

尝试以下操作:

有一个循环来计算有多少实际匹配项:

int num_matches = 0;
for( int idx = 0; idx < matchesMask.size(); idx++ )
{
    if ( matchesMask[idx] == 1 )
        num_matches++;
}

现在声明正确大小的 CvMats:

matched_points1  = cvCreateMat(2,numPoints,CV_32F);
matched_points2  = cvCreateMat(2,numPoints,CV_32F);

现在遍历 filteredMatches 并插入:(确切的语法可能有所不同，您明白了)

offset = 0;
for (int idx = 0; idx < matchesMask.size(); idx++)
{
    if ( matchesMask[idx] == 1 ) {
        matched_points1[2*offset] = keypoints1[idx].pt.x;
        matched_points1[2*offset+1] = keypoints1[idx].pt.y;
        matched_points2[2*offset] = keypoints2[filteredMatches[idx]].pt.x;
        matched_points2[2*offset+1] = keypoints2[filteredMatches[idx]].pt.y;
        offset++;
    }
}