img=imread('C:\Users\Shalaw\Downloads\gray.jpeg');
[n,m]=size(img);
part1=img(1:n/2,1:m/2);
part2=img(1:n/2,(m/2)+1:end);
part3=img(n/2+1:end,1:m/2);
part4=img(n/2+1:end,m/2+1:end);
for i=1:4
subplot(2,2,i);
imshow(part1); -----> I want to change 1 to i, but i cant please help.
end

1 comentario

Stephen23
Stephen23 el 15 de Jul. de 2019
Editada: Stephen23 el 15 de Jul. de 2019
Simply use a cell array with indexing:
Putting numbers into variable names is a sign that you are doing something wrong. In most cases using indexing is simpler, neater, and much more efficient.

Iniciar sesión para comentar.

 Respuesta aceptada

Adam
Adam el 15 de Jul. de 2019

0 votos

img=imread('C:\Users\Shalaw\Downloads\gray.jpeg');
[n,m]=size(img);
part{1}=img(1:n/2,1:m/2);
part{2}=img(1:n/2,(m/2)+1:end);
part{3}=img(n/2+1:end,1:m/2);
part{4}=img(n/2+1:end,m/2+1:end);
for i=1:4
subplot(2,2,i);
imshow(part{i});
end

Más respuestas (1)

Andrei Bobrov
Andrei Bobrov el 15 de Jul. de 2019

0 votos

img=imread('C:\Users\Shalaw\Downloads\gray.jpeg');
[n,m]=size(img);
imgs = mat2cell(img,[n,n]/2,[m,m]/2)';
for ii = 1:4
subplot(2,2,ii);
imshow(imgs{ii});
end

3 comentarios

shalaw faraj
shalaw faraj el 15 de Jul. de 2019
your answre is greate, what about 4*4 and 8*8 and .... ?
very thanks.
Andrei Bobrov
Andrei Bobrov el 15 de Jul. de 2019
k = 4;
[n,m]=size(img);
imgs = mat2cell(img, n/k*ones(1,k), m/k*ones(1,k))';
for ii = 1:k^2
subplot(k,k,ii);
imshow(imgs{ii});
end
shalaw faraj
shalaw faraj el 15 de Jul. de 2019
I am very sorry, but I want to find (mean, max, min, std, entropy) for each block and using for but it is not working please help. thanks again.
img=imread('C:\Users\Shalaw\Downloads\gray.jpeg');
[n,m]=size(img);
part{1}=img(1:n/2,1:m/2);
part{2}=img(1:n/2,(m/2)+1:end);
part{3}=img(n/2+1:end,1:m/2);
part{4}=img(n/2+1:end,m/2+1:end);
for i=1:4
subplot(2,2,i);
imshow(part{i});
Xmean{i}=mean(part{i});
Xmax{i}=max(part{i});
Xmin{i}=min(part{i});
Xstd{i}=std(part{i});
Xentropy{i}=entropy(part{i});
end

Iniciar sesión para comentar.

Categorías

Preguntada:

el 15 de Jul. de 2019

Comentada:

el 15 de Jul. de 2019

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by