如何匹配SIFT中的关键点？
我为图像中的每个关键点计算了128个大小的矢量。
让I1是原始图像，I2是45度旋转图像。
我获得了I1的130个关键点和I2的104个关键点。
即。 128x130和128x104。
我计算了I1的一个关键点与I2的所有关键点之间的欧几里德距离。所以我得到了尺寸为128x104的欧几里德距离矩阵。
现在我需要从这个欧氏距离矩阵中选择最近的关键点。如何从128 x 104大小的矩阵中选择最小距离128大小的矢量？
答案 0 :(得分：1)
由于你已经计算了关键点之间的距离，为了匹配它们，按欧几里德距离的递增顺序对它们进行排序，并且只考虑那些常数* min_distance的关键点[即：在某些％年龄选择已排序的距离]为'良好匹配'。
OpenCV中还有BruteForceMatcher，KNNMatch和FlannBasedMatcher（下面的URL） http://docs.opencv.org/2.4/doc/tutorials/features2d/feature_flann_matcher/feature_flann_matcher.html#feature-flann-matcher
和
另外，请看看这些问题及其回答。
1）Trying to match two images using sift in OpenCv, but too many matches
2）Efficient way for SIFT descriptor matching
为了完整起见，提供一些非常粗略的代码供您参考。
If you have ;
class SIFTDemo
{
private:
Mat image;
vector<cv::KeyPoint> keypoints;
Mat descriptors;
Mat sift_output;
vector<DMatch> matches;
public:
SIFTDemo();
~SIFTDemo();
SIFTDemo(Mat m);
void extractSiftFeatures();
vector <DMatch> FindMatchesEuclidian(SIFTDemo &m2);
};
Then one can have something like this;
void SIFTDemo::extractSiftFeatures()
{
SIFT siftobject;
siftobject.operator()(image, Mat(), keypoints, descriptors);
}
vector<DMatch> SIFTDemo::FindMatchesEuclidian(SIFTDemo &m2)
{
// Calculate euclidian distance between keypoints to find best matching pairs.
// create two dimensional vector for storing euclidian distance
vector< vector<float> > vec1, unsortedvec1;
for (int i=0; i<this->keypoints.size(); i++)
{
vec1.push_back(vector<float>()); // Add an empty row
unsortedvec1.push_back(vector<float>());
}
// create vector of DMatch for storing matxhes point
vector<DMatch> matches1;
DMatch dm1;
// loop through keypoints1.size
for (int i=0; i<this->keypoints.size(); i++)
{
// get 128 dimensions in a vector
vector<float> k1;
for(int x=0; x<128; x++)
{
k1.push_back((float)this->descriptors.at<float>(i,x));
}
// loop through keypoints2.size
for (int j=0; j<m2.keypoints.size(); j++)
{
double temp=0;
// calculate euclidian distance
for(int x=0; x<128; x++)
{
temp += (pow((k1[x] - (float)m2.descriptors.at<float>(j,x)), 2.0));
}
vec1[i].push_back((float)sqrt(temp)); // store distance for each keypoints in image2
unsortedvec1[i] = vec1[i];
}
sort(vec1[i].begin(),vec1[i].end()); // sort the vector distances to get shortest distance
// find position of the shortest distance
int pos = (int)(find(unsortedvec1[i].begin(), unsortedvec1[i].end(), vec1[i][0]) - unsortedvec1[i].begin());
// assign that matchin feature to DMatch variable dm1
dm1.queryIdx = i;
dm1.trainIdx = pos;
dm1.distance = vec1[i][0];
matches1.push_back(dm1);
this->matches.push_back(dm1);
//cout << pos << endl;
}
// craete two dimensional vector for storing euclidian distance
vector<vector<float>> vec2, unsortedvec2;
for (int i=0; i<m2.keypoints.size(); i++)
{
vec2.push_back(vector<float>()); // Add an empty row
unsortedvec2.push_back(vector<float>());
}
// create vector of DMatch for storing matxhes point
vector<DMatch> matches2;
DMatch dm2;
// loop through keypoints2.size
for (int i=0; i<m2.keypoints.size(); i++)
{
// get 128 dimensions in a vector
vector<float> k1;
for(int x=0; x<128; x++)
{
k1.push_back((float)m2.descriptors.at<float>(i,x));
}
// loop through keypoints1.size
for (int j=0; j<this->keypoints.size(); j++)
{
double temp=0;
// calculate euclidian distance
for(int x=0; x<128; x++)
{
temp += (pow((k1[x] - (float)this->descriptors.at<float>(j,x)), 2.0));
}
vec2[i].push_back((float)sqrt(temp)); // store distance for each keypoints in image1
unsortedvec2[i] = vec2[i];
}
sort(vec2[i].begin(),vec2[i].end()); // sort the vector distances to get shortest distance
// find position of the shortest distance
int pos = (int)(find(unsortedvec2[i].begin(), unsortedvec2[i].end(), vec2[i][0]) - unsortedvec2[i].begin());
// assign that matchin feature to DMatch variable
dm2.queryIdx = i;
dm2.trainIdx = pos;
dm2.distance = vec2[i][0];
matches2.push_back(dm2);
m2.matches.push_back(dm2);
//cout << pos << endl;
}
// Ref : http://docs.opencv.org/2.4/doc/tutorials/features2d/feature_flann_matcher/feature_flann_matcher.html#feature-flann-matcher
//-- Quick calculation of max and min distances between keypoints1
double max_dist = 0;
double min_dist = 500.0;
for( int i = 0; i < matches1.size(); i++ )
{
double dist = matches1[i].distance;
if( dist < min_dist ) min_dist = dist;
if( dist > max_dist ) max_dist = dist;
}
// Draw only "good" matches1 (i.e. whose distance is less than 2*min_dist )
vector<DMatch> good_matches1;
for( int i = 0; i < matches1.size(); i++ )
{
if( matches1[i].distance <= 2*min_dist )
{
good_matches1.push_back( matches1[i]);
}
}
// Quick calculation of max and min distances between keypoints2 but not used
for( int i = 0; i < matches2.size(); i++ )
{
double dist = matches2[i].distance;
if( dist < min_dist ) min_dist = dist;
if( dist > max_dist ) max_dist = dist;
}
// Draw only "good" matches by comparing that (ft1 gives ft2) and (ft2 gives ft1)
vector<DMatch> good_matches;
for(unsigned int i=0; i<good_matches1.size(); i++)
{
// check ft1=ft2 and ft2=ft1
if(good_matches1[i].queryIdx == matches2[good_matches1[i].trainIdx].trainIdx)
good_matches.push_back(good_matches1[i]);
}
return good_matches;
}
最后，正如评论中所提到的，还要看RANSAC这样做。不要深入研究，不要让答案更长，但你可以在网上和SO上找到资源。