我想匹配立体图像中的要素点。我已经用不同的算法找到并提取了特征点，现在我需要一个很好的匹配。在这种情况下，我使用FAST算法进行检测和提取，使用BruteForceMatcher来匹配要素点。
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]);
}
因为有很多错误的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，我要么得到很多假匹配，要么只有几个正确的匹配（见下图）。
many matches with bad results http://codemax.de/upl/badMatchesFAST.png only a few good matches http://codemax.de/upl/goodMatchesFAST.png
我认为这不是编程问题，而是更匹配的问题。据我所知，BruteForceMatcher仅考虑特征点的视觉距离（存储在FeatureExtractor中），而不是局部距离（x＆amp; y位置），这在我的情况下很重要，太。有没有人有这个问题的经验或改善匹配结果的好主意？
修改
我改变了代码，它给了我50个最佳匹配。在此之后，我通过第一场比赛检查，是否在指定区域。如果不是，我接下一场比赛，直到我在给定区域内找到一场比赛。
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比赛，但有了这个我能够删除更多的错误匹配：
答案 0 :(得分：27)
确定高质量特征匹配的另一种方法是David Lowe在SIFT的论文中提出的比率测试（第20页的解释）。该测试通过计算最佳和次佳匹配之间的比率来拒绝不良匹配。如果比率低于某个阈值，则匹配将被丢弃为低质量。
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]);
}
}
答案 1 :(得分：23)
通过比较所有特征检测算法，我发现了一个很好的组合，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，匹配是否在定义的区域内我得到了很多匹配，请看图像：
a lot of good matches with FAST and SIFT http://codemax.de/upl/FASTandSIFT.png
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();
}
}
}
答案 2 :(得分：7)
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;
}
}
}
}
因为它的立体图像使用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);
}
}
}