将Matlab高斯导数转换为Opencv
我试图将我在matlab中制作的旧练习转换为OpenCV。代码发布在下面。我还没有能够在OpenCv中找到任何能够实现我想要的功能，可能是因为其他名称然后我所期待的。
以下是每个位置的最大响应作为标签时的输出。显然有些事情很糟糕。
这是matlab代码：
function responses = getBifResponsesEx(im, myEps, sigma, kernelSize)
if ( nargin == 3 )
if ( sigma >= 1 )
kernelSize = 6*sigma + 1;
else
kernelSize = 7;
end
end
responses = zeros(size(im,1), size(im,2), 7);
%
% Gaussian derivatives
%
kernVal = ceil(kernelSize/2) - 1;
x = (-kernVal:kernVal);
g = 1/(2*pi*sigma^2)*exp(-(x.^2./(2*(sigma^2))));
g = g/sum(g);
dg = -2*x/(2*sigma^2).*g*sigma;
ddg = ((2*x/(2*sigma^2)).^2 - 1/(sigma^2)).*g*sigma;
%
% Gaussian convolution of the image
%
s00 = filter2(g, im);
s00 = filter2(g', s00);
s10 = filter2(g', im);
s10 = filter2(dg, s10);
s01 = filter2(g, im);
s01 = filter2(dg', s01);
s11 = filter2(dg, im);
s11 = filter2(dg', s11);
s20 = filter2(g', im);
s20 = filter2(g', s20);
s20 = filter2(ddg, s20);
s02 = filter2(g, im);
s02 = filter2(g, s02);
s02 = filter2(ddg', s02);
%
% Symmetry types - MISSING CODE!!!!
%
lam = sigma^2*(s20+s02);
gam = sigma^2*(sqrt((s20-s02).^2+4*s11.^2));
responses(:,:,1) = myEps*s00;
responses(:,:,2) = 2*sigma*sqrt(s10.^2+s01.^2);
responses(:,:,3) = +lam;
responses(:,:,4) = -lam;
responses(:,:,5) = 2^-.5*(gam+lam);
responses(:,:,6) = 2^-.5*(gam-lam);
responses(:,:,7) = gam;
end
这是我的转换页面。从我所看到的情况来看，它对s20，s02的反应很不满意。任何人都能告诉我该怎么做？
void extract_bif_features(const cv::Mat & src,
std::vector<cv::Mat> & dst, BIFParams params)
{
float sigma = params.sigma;
float n=0;
int kernelSize;
if(sigma>=1)
kernelSize = 6*sigma + 1;
else
kernelSize = 7;
cv::Mat gray,p00,p10,p01,p11,p20,p02;
cv::cvtColor(src,gray,CV_BGR2GRAY);
auto kernVal = (int)ceil(kernelSize/2.0) - 1;
cv::Mat_<float> g(1,kernelSize);float*gp = g.ptr<float>();
cv::Mat_<float> dg(1,kernelSize);float*dgp = dg.ptr<float>();
cv::Mat_<float> ddg(1,kernelSize); float*ddgp = ddg.ptr<float>();
cv::Mat_<float> X(1,kernelSize);float*xp = X.ptr<float>();
auto gsum=0.0f;
for(int x = -kernVal;x<=kernVal;++x)
{
xp[x+kernVal] = x;
gp[x+kernVal] = 1/(2*CV_PI*sigma*sigma)*exp(-(x*x/(2*(sigma*sigma))));
gsum += gp[x+kernVal];
}
g = g/gsum;
cv::multiply((-2*X / (2*sigma*sigma)),g*sigma,dg);
cv::pow((2*X/(2*sigma*sigma)),2,ddg);
ddg -=1/(sigma*sigma);
cv::multiply(ddg,g*sigma,ddg);
std::cout << ddg<< std::endl;
std::cout << dg<< std::endl;
cv::sepFilter2D(gray,p00,CV_32FC1,g,g,cv::Point(-1,-1),0.0,cv::BORDER_REPLICATE); cv::sepFilter2D(gray,p01,CV_32FC1,dg,g,cv::Point(-1,-1),0.0,cv::BORDER_REPLICATE); cv::sepFilter2D(gray,p10,CV_32FC1,g,dg,cv::Point(-1,-1),0.0,cv::BORDER_REPLICATE); cv::sepFilter2D(gray,p11,CV_32FC1,dg,dg,cv::Point(-1,-1),0.0,cv::BORDER_REPLICATE);
//NOT SURE HERE
cv::sepFilter2D(gray,p20,CV_32FC1,g,ddg,cv::Point(-1,-1),0.0,cv::BORDER_REPLICATE);
//cv::sepFilter2D(p20,p20,CV_32FC1,1,g,cv::Point(-1,-1),0.0,cv::BORDER_REPLICATE); cv::sepFilter2D(gray,p02,CV_32FC1,g,ddg,cv::Point(-1,-1),0.0,cv::BORDER_REPLICATE); //cv::sepFilter2D(p02,p02,CV_32FC1,g,1,cv::Point(-1,-1),0.0,cv::BORDER_REPLICATE);
cv::filter2D(gray,p20,CV_32FC1,g,cv::Point(-1,-1),0.0,cv::BORDER_REPLICATE);
cv::filter2D(p20,p20,CV_32FC1,g,cv::Point(-1,-1),0.0,cv::BORDER_REPLICATE);
cv::filter2D(p20,p20,CV_32FC1,ddg,cv::Point(-1,-1),0.0,cv::BORDER_REPLICATE);
cv::filter2D(gray,p02,CV_32FC1,g,cv::Point(-1,-1),0.0,cv::BORDER_REPLICATE);
cv::filter2D(p02,p02,CV_32FC1,g.t(),cv::Point(-1,-1),0.0,cv::BORDER_REPLICATE);
cv::filter2D(p02,p02,CV_32FC1,ddg.t(),cv::Point(-1,-1),0.0,cv::BORDER_REPLICATE);
dst.resize(6);
auto sigma_square = sigma*sigma;
cv::Mat Lam = sigma_square * (p20+p02);
cv::Mat Gam ;
cv::sqrt((((p20-p02)*(p20-p02))+4*p11*p11),Gam);
Gam *= sigma_square ;
cv::Mat test = p10*p10;
//slop
cv::sqrt(p10*p10 + p01*p01,dst[0]);
dst[0] = dst[0]*2*sigma;//slop
//blob
dst[1] = Lam;
dst[2] = -1*Lam;
//line
dst[3] = sqrt(2.0f)*(Gam+Lam);
dst[4] = sqrt(2.0f)*(Gam-Lam);
//saddle
dst[5] = Gam;
}
2 个答案:
答案 0 :(得分：1)
答案来自于我得到的信息。
CV ::乘（（P20-P02），（P20-P02），GAM）;与Gam =（p20-p02）*（p20-p02）;
不同完整代码：根据最高回应分类，Griffin（2008）。
RIDAR_API void extract_bif_features(const cv::Mat & src,
std::vector<cv::Mat> & dst, BIFParams params)
{
float sigma = params.sigma;
float eta = params.eta;
int kernelSize;
if(sigma>=1)
kernelSize = 4*sigma + 1;
else
kernelSize = 5;
auto kernVal = (int)ceil(kernelSize/2.0) - 1;
cv::Mat_<float> g(1,kernelSize);float*gp = g.ptr<float>();
cv::Mat_<float> X(1,kernelSize);float*xp = X.ptr<float>();
auto gsum=0.0f;
for(int x = -kernVal;x<=kernVal;++x)
{
xp[x+kernVal] = x;
gp[x+kernVal] = 1/(2*CV_PI*sigma*sigma)*exp(-(x*x/(2*(sigma*sigma))));
gsum += gp[x+kernVal];
} g = g/gsum;
cv::Mat dg = -2*X.mul(g*sigma) / (2*sigma*sigma);
cv::Mat ddg = ((2*X/(2*sigma*sigma)).mul((2*X/(2*sigma*sigma))) - 1/(sigma*sigma)).mul(g*sigma);
cv::Mat gray,p00,p10,p01,p11,p20,p02;
cv::cvtColor(src,gray,CV_BGR2GRAY);
cv::filter2D(gray,p00,CV_32FC1,g);
cv::filter2D(p00,p00,CV_32FC1,g.t());
cv::filter2D(gray,p10,CV_32FC1,g.t());
cv::filter2D(p10,p10,CV_32FC1,dg);
cv::filter2D(gray,p01,CV_32FC1,g);
cv::filter2D(p01,p01,CV_32FC1,dg.t());
cv::filter2D(gray,p11,CV_32FC1,dg);
cv::filter2D(p11,p11,CV_32FC1,dg.t());
cv::filter2D(gray,p20,CV_32FC1,g.t());
cv::filter2D(p20,p20,CV_32FC1,g.t());
cv::filter2D(p20,p20,CV_32FC1,ddg);
cv::filter2D(gray,p02,CV_32FC1,g);
cv::filter2D(p02,p02,CV_32FC1,g);
cv::filter2D(p02,p02,CV_32FC1,ddg.t());
#ifdef DISPLAY_WHILE_RUNNING
double max,min;
cv::imshow("p00",p00/255);
//
cv::minMaxIdx(p01,&min,&max);
cv::imshow("p01",(p01-min)/(max-min));
//
cv::minMaxIdx(p10,&min,&max);
cv::imshow("p10",(p10-min)/(max-min));
cv::minMaxIdx(p11,&min,&max);
cv::imshow("p11",(p11-min)/(max-min));
cv::minMaxIdx(p02,&min,&max);
cv::imshow("p02",(p02-min)/(max-min));
cv::minMaxIdx(p20,&min,&max);
cv::imshow("p20",(p20-min)/(max-min));
cv::waitKey();
#endif
dst.resize(7);
auto sigma_square = sigma*sigma;
auto p2d = p20-p02;
//LAM
dst[2] = sigma_square * (p20+p02);
//GAM
cv::sqrt( (p2d).mul(p2d) + (4.0f * p11.mul(p11)) ,dst[6] );
dst[6] = dst[6] * sigma_square;
//FLAT
dst[0] = eta*p00;
//slop
cv::sqrt(p10.mul(p10)+p01.mul(p01),dst[1]);
dst[1] *= 2.0f*sigma;
//blob dst[2]
dst[3] = -dst[2];
//line
dst[4] = pow(2.0,-0.5)*(dst[6]+dst[2]);
dst[5] = pow(2.0,-0.5)*(dst[6]-dst[2]);
//saddle dst[6]
#ifdef DISPLAY_WHILE_RUNNING
double max,min;
cv::minMaxIdx(dst[0],&min,&max);//
cv::imshow("FLAT",(dst[0]-min)/(max-min));
cv::minMaxIdx(dst[1],&min,&max);//
cv::imshow("SLOPE",(dst[1]-min)/(max-min));
cv::minMaxIdx(dst[2],&min,&max);
cv::imshow("BLOB+",(dst[2]-min)/(max-min));
cv::minMaxIdx(dst[3],&min,&max);//
cv::imshow("BLOB-",(dst[3]-min)/(max-min));
cv::minMaxIdx(dst[4],&min,&max);//
cv::imshow("LINE+",(dst[4]-min)/(max-min));
cv::minMaxIdx(dst[5],&min,&max);
cv::imshow("LINE-",(dst[5]-min)/(max-min));
cv::minMaxIdx(dst[6],&min,&max);
cv::imshow("SADDLE",(dst[6]-min)/(max-min));
cv::waitKey();
#endif
}
答案 1 :(得分：0)
为什么不对dg和ddg采取平均值？
cv::filter2D(gray,p20,CV_32FC1,g.t());
cv::filter2D(p20,p20,CV_32FC1,g.t());
为什么要在这里进行两次过滤？
// GAM
cv :: sqrt（（p2d）.mul（p2d）+（4.0f * p11.mul（p11）），dst [6]）; dst [6] = dst [6] * sigma_square; 你得到这个公式的地方？
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