30

我想匹配立体图像中的特征点。我已经用不同的算法找到并提取了特征点，现在我需要一个好的匹配。在这种情况下，我使用 FAST 算法进行检测和提取以及BruteForceMatcher匹配特征点。

匹配代码：

vector< vector<DMatch> > matches;
//using either FLANN or BruteForce
Ptr<DescriptorMatcher> matcher = DescriptorMatcher::create(algorithmName);
matcher->knnMatch( descriptors_1, descriptors_2, matches, 1 );

//just some temporarily code to have the right data structure
vector< DMatch > good_matches2;
good_matches2.reserve(matches.size());  
for (size_t i = 0; i < matches.size(); ++i)
{ 
    good_matches2.push_back(matches[i][0]);     
}

因为有很多错误匹配，我计算了最小和最大距离并删除了所有太糟糕的匹配：

//calculation of max and min distances between keypoints
double max_dist = 0; double min_dist = 100;
for( int i = 0; i < descriptors_1.rows; i++ )
{
    double dist = good_matches2[i].distance;
    if( dist < min_dist ) min_dist = dist;
    if( dist > max_dist ) max_dist = dist;
}

//find the "good" matches
vector< DMatch > good_matches;
for( int i = 0; i < descriptors_1.rows; i++ )
{
    if( good_matches2[i].distance <= 5*min_dist )
    {
        good_matches.push_back( good_matches2[i]); 
    }
}

问题是，我要么得到很多错误匹配，要么只有少数正确匹配（见下图）。

许多比赛结果不佳
（来源：codemax.de）

只有几场不错的比赛
（来源：codemax.de）

我认为这不是编程问题，而是匹配的问题。据我了解，BruteForceMatcher仅考虑特征点的视觉距离（存储在 中FeatureExtractor），而不是局部距离（x&y 位置），这在我的情况下也很重要。有没有人遇到过这个问题或改善匹配结果的好主意？

编辑

我更改了代码，它为我提供了 50 个最佳匹配项。在此之后，我通过第一场比赛来检查它是否在指定区域。如果不是，我会进行下一场比赛，直到在给定区域内找到比赛为止。

vector< vector<DMatch> > matches;
Ptr<DescriptorMatcher> matcher = DescriptorMatcher::create(algorithmName);
matcher->knnMatch( descriptors_1, descriptors_2, matches, 50 );

//look if the match is inside a defined area of the image
double tresholdDist = 0.25 * sqrt(double(leftImageGrey.size().height*leftImageGrey.size().height + leftImageGrey.size().width*leftImageGrey.size().width));

vector< DMatch > good_matches2;
good_matches2.reserve(matches.size());  
for (size_t i = 0; i < matches.size(); ++i)
{ 
    for (int j = 0; j < matches[i].size(); j++)
    {
    //calculate local distance for each possible match
    Point2f from = keypoints_1[matches[i][j].queryIdx].pt;
    Point2f to = keypoints_2[matches[i][j].trainIdx].pt;        
    double dist = sqrt((from.x - to.x) * (from.x - to.x) + (from.y - to.y) * (from.y - to.y));
    //save as best match if local distance is in specified area
    if (dist < tresholdDist)
    {
        good_matches2.push_back(matches[i][j]);
        j = matches[i].size();
    }
}

我想我没有得到更多的匹配，但是有了这个我可以删除更多的错误匹配：

更少但更好的功能
（来源：codemax.de）

4

3 回答 3

33

确定高质量特征匹配的另一种方法是 David Lowe 在他关于SIFT的论文中提出的比率测试（第 20 页有解释）。此测试通过计算最佳匹配和次佳匹配之间的比率来拒绝不良匹配。如果该比率低于某个阈值，则匹配被丢弃为低质量。

std::vector<std::vector<cv::DMatch>> matches;
cv::BFMatcher matcher;
matcher.knnMatch(descriptors_1, descriptors_2, matches, 2);  // Find two nearest matches
vector<cv::DMatch> good_matches;
for (int i = 0; i < matches.size(); ++i)
{
    const float ratio = 0.8; // As in Lowe's paper; can be tuned
    if (matches[i][0].distance < ratio * matches[i][1].distance)
    {
        good_matches.push_back(matches[i][0]);
    }
}
于 2013-07-31T17:37:42.853 回答
25

通过比较所有的特征检测算法，我找到了一个很好的组合，这给了我更多的匹配。现在我使用 FAST 进行特征检测，使用 SIFT 进行特征提取，使用 BruteForce 进行匹配。结合检查，匹配项是否在定义的区域内，我得到了很多匹配项，见图：

与 FAST 和 SIFT 有很多很好的匹配
（来源：codemax.de）

相关代码：

Ptr<FeatureDetector> detector;
detector = new DynamicAdaptedFeatureDetector ( new FastAdjuster(10,true), 5000, 10000, 10);
detector->detect(leftImageGrey, keypoints_1);
detector->detect(rightImageGrey, keypoints_2);

Ptr<DescriptorExtractor> extractor = DescriptorExtractor::create("SIFT");
extractor->compute( leftImageGrey, keypoints_1, descriptors_1 );
extractor->compute( rightImageGrey, keypoints_2, descriptors_2 );

vector< vector<DMatch> > matches;
Ptr<DescriptorMatcher> matcher = DescriptorMatcher::create("BruteForce");
matcher->knnMatch( descriptors_1, descriptors_2, matches, 500 );

//look whether the match is inside a defined area of the image
//only 25% of maximum of possible distance
double tresholdDist = 0.25 * sqrt(double(leftImageGrey.size().height*leftImageGrey.size().height + leftImageGrey.size().width*leftImageGrey.size().width));

vector< DMatch > good_matches2;
good_matches2.reserve(matches.size());  
for (size_t i = 0; i < matches.size(); ++i)
{ 
    for (int j = 0; j < matches[i].size(); j++)
    {
        Point2f from = keypoints_1[matches[i][j].queryIdx].pt;
        Point2f to = keypoints_2[matches[i][j].trainIdx].pt;

        //calculate local distance for each possible match
        double dist = sqrt((from.x - to.x) * (from.x - to.x) + (from.y - to.y) * (from.y - to.y));

        //save as best match if local distance is in specified area and on same height
        if (dist < tresholdDist && abs(from.y-to.y)<5)
        {
            good_matches2.push_back(matches[i][j]);
            j = matches[i].size();
        }
    }
}
于 2013-08-02T15:27:53.507 回答
9

除了比率测试，您还可以：

仅使用对称匹配：

void symmetryTest(const std::vector<cv::DMatch> &matches1,const std::vector<cv::DMatch> &matches2,std::vector<cv::DMatch>& symMatches)
{
    symMatches.clear();
    for (vector<DMatch>::const_iterator matchIterator1= matches1.begin();matchIterator1!= matches1.end(); ++matchIterator1)
    {
        for (vector<DMatch>::const_iterator matchIterator2= matches2.begin();matchIterator2!= matches2.end();++matchIterator2)
        {
            if ((*matchIterator1).queryIdx ==(*matchIterator2).trainIdx &&(*matchIterator2).queryIdx ==(*matchIterator1).trainIdx)
            {
                symMatches.push_back(DMatch((*matchIterator1).queryIdx,(*matchIterator1).trainIdx,(*matchIterator1).distance));
                break;
            }
        }
    }
}

并且由于它是立体图像，因此使用 ransac 测试：

void ransacTest(const std::vector<cv::DMatch> matches,const std::vector<cv::KeyPoint>&keypoints1,const std::vector<cv::KeyPoint>& keypoints2,std::vector<cv::DMatch>& goodMatches,double distance,double confidence,double minInlierRatio)
{
    goodMatches.clear();
    // Convert keypoints into Point2f
    std::vector<cv::Point2f> points1, points2;
    for (std::vector<cv::DMatch>::const_iterator it= matches.begin();it!= matches.end(); ++it)
    {
        // Get the position of left keypoints
        float x= keypoints1[it->queryIdx].pt.x;
        float y= keypoints1[it->queryIdx].pt.y;
        points1.push_back(cv::Point2f(x,y));
        // Get the position of right keypoints
        x= keypoints2[it->trainIdx].pt.x;
        y= keypoints2[it->trainIdx].pt.y;
        points2.push_back(cv::Point2f(x,y));
    }
    // Compute F matrix using RANSAC
    std::vector<uchar> inliers(points1.size(),0);
    cv::Mat fundemental= cv::findFundamentalMat(cv::Mat(points1),cv::Mat(points2),inliers,CV_FM_RANSAC,distance,confidence); // confidence probability
    // extract the surviving (inliers) matches
    std::vector<uchar>::const_iterator
    itIn= inliers.begin();
    std::vector<cv::DMatch>::const_iterator
    itM= matches.begin();
    // for all matches
    for ( ;itIn!= inliers.end(); ++itIn, ++itM)
    {
        if (*itIn)
        { // it is a valid match
            goodMatches.push_back(*itM);
        }
    }
}
于 2013-08-04T02:33:00.140 回答