Azzera filtri
Azzera filtri

An Edge-Guided Image Interpolation Algorithm via Directional Filtering and Data Fusion

1 visualizzazione (ultimi 30 giorni)
function A=esintp(I)
%%%%%process A%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[n,m]=size(I);
n=2*n;m=2*m;
A=zeros(n,m);
A(1:2:n,1:2:m)=I;
Ax=A;Ay=A;
f=[-1/16 9/16 9/16 -1/16];
for i=4:2:n-4
for j=4:2:m-4
x=[A(i-3,j-3) A(i-1,j-1) A(i+1,j+1) A(i+3,j+3)];
y=[A(i+3,j-3) A(i+1,j-1) A(i-1,j+1) A(i-3,j+3)];
Ax(i,j)=sum(x.*f);
Ay(i,j)=sum(f.*y);
end
end
for i=4:2:n-4
for j=4:2:m-4
zx=[Ax(i-2,j-2) Ax(i-1,j-1) Ax(i,j) Ax(i+1,j+1) Ax(i+2,j+2)];
zy=[Ay(i+2,j-2) Ay(i+1,j-1) Ay(i,j) Ay(i-1,j+1) Ay(i-2,j+2)];
x=[A(i-1,j-1) A(i+1,j+1) A(i+1,j-1) A(i-1,j+1)];
mx=mean(x);
px=sum((x-mx).^2)/3+0.1;
pzx=sum((zx-mx).^2)/4+0.1;
pzy=sum((zy-mx).^2)/4+0.1;
pv1=pzx-px;pv1=max(0.01,pv1);
pv2=pzy-px;pv2=max(0.01,pv2);
R=[pv1 0;0 pv2];
y=[Ax(i,j) Ay(i,j)]';
H=[1;1];
A(i,j)=mx+px*H'*(H*px*H'+R)^(-1)*(y-H*mx);
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Ax=A;Ay=A;
f=[-1/16 0 9/16 1 9/16 0 -1/16];
Ax=conv2(Ax,f);
Ax=Ax(:,4:m+3);
Ay=conv2(Ay,f');
Ay=Ay(4:n+3,:);
c=1.5;
h=[-1 5 5 -1];h=h/sum(h);
for i=5:2:n-5
for j=4:2:m-4
y=[Ax(i-2,j) A(i-1,j) A(i+1,j) Ax(i+2,j)];
my=sum(h.*y);
x=[Ax(i,j-2) Ax(i,j-1) Ax(i,j) Ax(i,j+1) Ax(i,j+2)];
vx=cov(x)+0.1;
vy=c*cov([y my])+0.1;
A(i,j)=(Ax(i,j)*vy+my*vx)/(vx+vy);
end
end
for i=4:2:n-4
for j=5:2:m-5
x=[Ay(i,j-2) A(i,j-1) A(i,j+1) Ay(i,j+2)];
mx=sum(h.*x);
y=[Ay(i-2,j) Ay(i-1,j) Ay(i,j) Ay(i+1,j) Ay(i+2,j)];
vx=c*cov([x mx])+0.1;
vy=cov(y)+0.1;
A(i,j)=(mx*vy+Ay(i,j)*vx)/(vx+vy);
end
end
return;
can anyone explain this code???

Risposte (2)

Image Analyst
Image Analyst il 7 Giu 2014
I bet the author could, but probably doesn't want to, or else he or she would have done so by adding comments. I'm not even going to attempt it.

qing zhao
qing zhao il 13 Nov 2017
Modificato: qing zhao il 13 Nov 2017
Can someone provide some suggestions about above code.

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