[英]how to find euclidean distance between pixels within a image in opencv
[英]How to find euclidean distance between keypoints of a single image in opencv
我想为图像中的每个关键点获取距离矢量d。 距离矢量应包括从该关键点到该图像中所有其他关键点的距离。 注意:使用SIFT找到关键点。
我是opencv的新手。 C ++中是否有库函数可以让我的任务变得简单?
如果你对位置距离感兴趣,但描述符距离你可以使用这个:
cv::Mat SelfDescriptorDistances(cv::Mat descr)
{
cv::Mat selfDistances = cv::Mat::zeros(descr.rows,descr.rows, CV_64FC1);
for(int keyptNr = 0; keyptNr < descr.rows; ++keyptNr)
{
for(int keyptNr2 = 0; keyptNr2 < descr.rows; ++keyptNr2)
{
double euclideanDistance = 0;
for(int descrDim = 0; descrDim < descr.cols; ++descrDim)
{
double tmp = descr.at<float>(keyptNr,descrDim) - descr.at<float>(keyptNr2, descrDim);
euclideanDistance += tmp*tmp;
}
euclideanDistance = sqrt(euclideanDistance);
selfDistances.at<double>(keyptNr, keyptNr2) = euclideanDistance;
}
}
return selfDistances;
}
这将给你一个N×N矩阵(N =关键点的数量),其中Mat_i,j =关键点i和j之间的欧几里德距离。
有了这个输入:
我得到这些输出:
备注:由于距离是对称的,你可以在矩阵的计算中优化很多东西!
更新:
这是另一种方法:
从您的聊天中我知道您需要13GB内存来保存41381个关键点(您尝试过)的距离信息。 如果您只想要N个最佳匹配,请尝试以下代码:
// choose double here if you are worried about precision!
#define intermediatePrecision float
//#define intermediatePrecision double
//
void NBestMatches(cv::Mat descriptors1, cv::Mat descriptors2, unsigned int n, std::vector<std::vector<float> > & distances, std::vector<std::vector<int> > & indices)
{
// TODO: check whether descriptor dimensions and types are the same for both!
// clear vector
// get enough space to create n best matches
distances.clear();
distances.resize(descriptors1.rows);
indices.clear();
indices.resize(descriptors1.rows);
for(int i=0; i<descriptors1.rows; ++i)
{
// references to current elements:
std::vector<float> & cDistances = distances.at(i);
std::vector<int> & cIndices = indices.at(i);
// initialize:
cDistances.resize(n,FLT_MAX);
cIndices.resize(n,-1); // for -1 = "no match found"
// now find the 3 best matches for descriptor i:
for(int j=0; j<descriptors2.rows; ++j)
{
intermediatePrecision euclideanDistance = 0;
for( int dim = 0; dim < descriptors1.cols; ++dim)
{
intermediatePrecision tmp = descriptors1.at<float>(i,dim) - descriptors2.at<float>(j, dim);
euclideanDistance += tmp*tmp;
}
euclideanDistance = sqrt(euclideanDistance);
float tmpCurrentDist = euclideanDistance;
int tmpCurrentIndex = j;
// update current best n matches:
for(unsigned int k=0; k<n; ++k)
{
if(tmpCurrentDist < cDistances.at(k))
{
int tmpI2 = cIndices.at(k);
float tmpD2 = cDistances.at(k);
// update current k-th best match
cDistances.at(k) = tmpCurrentDist;
cIndices.at(k) = tmpCurrentIndex;
// previous k-th best should be better than k+1-th best //TODO: a simple memcpy would be faster I guess.
tmpCurrentDist = tmpD2;
tmpCurrentIndex =tmpI2;
}
}
}
}
}
它计算第一描述符与第二描述符的每个关键点的N个最佳匹配。 因此,如果您想为相同的关键点执行此操作,您将设置为descriptors1 = descriptors2
离线,如下所示。 记住:函数不知道两个描述符集是相同的,所以第一个最佳匹配(或至少一个)将是关键点本身,距离0总是! 如果使用结果,请记住这一点!
以下是生成类似于上图的示例代码:
int main()
{
cv::Mat input = cv::imread("../inputData/MultiLena.png");
cv::Mat gray;
cv::cvtColor(input, gray, CV_BGR2GRAY);
cv::SiftFeatureDetector detector( 7500 );
cv::SiftDescriptorExtractor describer;
std::vector<cv::KeyPoint> keypoints;
detector.detect( gray, keypoints );
// draw keypoints
cv::drawKeypoints(input,keypoints,input);
cv::Mat descriptors;
describer.compute(gray, keypoints, descriptors);
int n = 4;
std::vector<std::vector<float> > dists;
std::vector<std::vector<int> > indices;
// compute the N best matches between the descriptors and themselves.
// REMIND: ONE best match will always be the keypoint itself in this setting!
NBestMatches(descriptors, descriptors, n, dists, indices);
for(unsigned int i=0; i<dists.size(); ++i)
{
for(unsigned int j=0; j<dists.at(i).size(); ++j)
{
if(dists.at(i).at(j) < 0.05)
cv::line(input, keypoints[i].pt, keypoints[indices.at(i).at(j)].pt, cv::Scalar(255,255,255) );
}
}
cv::imshow("input", input);
cv::waitKey(0);
return 0;
}
keypoint
类有一个名为pt
的成员,后者又将x
和y
[点的(x,y)位置]作为自己的成员。
给定两个关键点kp1
和kp2
,然后很容易计算出欧氏距离:
Point diff = kp1.pt - kp2.pt;
float dist = std::sqrt( diff.x * diff.x + diff.y * diff.y )
在您的情况下,它将是一个迭代所有关键点的双循环。
std::vector< std::vector< float > > item;
按照a-Jays的建议计算距离
Point diff = kp1.pt - kp2.pt; float dist = std::sqrt( diff.x * diff.x + diff.y * diff.y );
使用push_back为每个关键点添加此向量 - > N次。
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