How to calculate horizontal and vertical correlation of an image in matlab???
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- How to calculate horizontal and vertical correlation of an image in matlab???
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Hyderkkk
am 26 Sep. 2021
IA=rgb2gray(im2double(orginal ));
c_diag = corrcoef(IA(1:end-1, 1:end-1), IA(2:end, 2:end));
c_vert = corrcoef(IA(1:end-1, :), IA(2:end, :));
c_horz = corrcoef(IA(:, 1:end-1, :), IA(:, 2:end));
figure;
subplot(3,3,1)
plot(IA(1:end-1, 1:end-1), IA(2:end, 2:end),'b'),title(sprintf('%s Horizontal Orginal image',b));
subplot(3,3,2)
plot(IA(1:end-1, :), IA(2:end, :),'b'),title(sprintf('%s Vertical Orginal image',b));
subplot(3,3,3)
plot(IA(:, 1:end-1, :), IA(:, 2:end),'b'),title(sprintf('%s Diagonal Orginal image',b));
IAA=rgb2gray(im2double(encrption));
c_diage = corrcoef(IAA(1:end-1, 1:end-1), IAA(2:end, 2:end))
c_verte = corrcoef(IAA(1:end-1, :), IAA(2:end, :))
c_horze = corrcoef(IAA(:, 1:end-1, :), IAA(:, 2:end))
subplot(3,3,4)
plot(IAA(1:end-1, 1:end-1), IAA(2:end, 2:end),'r'),title(sprintf('%s Horizontal chiphered image ',b));
subplot(3,3,5)
plot(IAA(1:end-1, :), IA(2:end, :),'r'),title(sprintf('%s Vertical chiphered image',b));
subplot(3,3,6)
plot(IAA(:, 1:end-1, :), IA(:, 2:end),'r'),title(sprintf('%s Diagonal chiphered image',b));
Antworten (3)
elhabib
am 30 Okt. 2017
% horizontal function rxy=H_correlation(image1)
P=double(image1);
x1 = double(P(:,1:end-1));
y1 = double(P(:,2:end));
randIndex1 = randperm(numel(x1));
randIndex1 = randIndex1(1:1000);
x = x1(randIndex1);
y = y1(randIndex1);
Ex=(1/(1000))*sum(x);
Ey=(1/(1000))*sum(y);
Dx=(1/(1000))*sum((x(:)-Ex).^2);
Dy=(1/(1000))*sum((y(:)-Ey).^2);
coxy=(1/(1000))*(sum((x-Ex).*(y-Ey)));
rxy=coxy/(sqrt(Dx*Dy));
% vertical
function Vxy=V_correlation(image1)
P=double(image1);
x1 = double(P(1:end-1,:));
y1 = double(P(2:end,:));
randIndex1 = randperm(numel(x1));
randIndex1 = randIndex1(1:1000);
x = x1(randIndex1);
y = y1(randIndex1);
Ex=(1/(1000))*sum(x);
Ey=(1/(1000))*sum(y);
Dx=(1/(1000))*sum((x(:)-Ex).^2);
Dy=(1/(1000))*sum((y(:)-Ey).^2);
coxy=(1/(1000))*(sum((x-Ex).*(y-Ey)));
Vxy=coxy/(sqrt(Dx*Dy));
3 Kommentare
Mahdiyar
am 10 Apr. 2015
Hi singhal
May you explain more about horizontal and vertical correlation?
Regards,
0 Kommentare
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