could you fix these codes?

3 vues (au cours des 30 derniers jours)
Tia
Tia le 14 Mai 2013
I=imread('filename');
a1=I(1:8,1:8);
a2=I(1:8,9:16);
a3=I(1:8,17:24);
a4=I(1:8,25:32);
b1=I(9:16,1:8);
b2=I(9:16,9:16);
b3=I(9:16,17:24);
b4=I(9:16,25:32);
c1=I(17:24,1:8);
c2=I(17:24,9:16);
c3=I(17:24,17:24);
c4=I(17:24,25:32);
d1=I(25:32,1:8);
d2=I(25:32,9:16);
d3=I(25:32,17:24);
d4=I(25:32,25:32);
could you tell me how to make it short? Thank you

Réponse acceptée

Jan
Jan le 14 Mai 2013
Modifié(e) : Jan le 14 Mai 2013
Don't do this.
Hiding the index in the name of a variable is a bad programming pattern. It is much more efficient to use indices as indices:
img = reshape(a(1:32, 1:32), 8, 4, 8, 4);
img = permute(img, [1,3,2,4]);
Now you have img(:, :, 1, 1) instead of a1.
This is much faster, needs less memory, and allows to expand the method for millions of tiles easily.
  3 commentaires
Image Analyst
Image Analyst le 16 Mai 2013
You may not even need to do that. It's possible that you do not need to even store the tiles at all. Do you? Why do these tiles need to be stored rather than just operated on and thrown away (reused)? Just operate on one tile at a time if you can - why store forever?
Tia
Tia le 15 Juil 2013
i try it. but it can't be back at 1D.. how do i get it? is it possible? or any ideas how to spilt it into 8x8pixels,but i want to manipulate each 8x8pixels,then store the tiles at all? thank you for your time

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Plus de réponses (2)

Iman Ansari
Iman Ansari le 14 Mai 2013
Modifié(e) : Iman Ansari le 14 Mai 2013
Hi. Use cell array:
I=imread('moon.tif');
C=mat2cell(I(1:32,1:32),8*ones(1,4),8*ones(1,4));
imshow(C{5},'InitialMagnification','fit')
C{5} or C{1,2} is a2 in your code.
  3 commentaires
Iman Ansari
Iman Ansari le 14 Mai 2013
It's not good having variable named a1,a2,..., see this:
I=imread('moon.tif');
C=mat2cell(I(1:32,1:32),8*ones(1,4),8*ones(1,4));
imshow(C{5},'InitialMagnification','fit')
C in this code is:
C={a1,a2,a3,a4;b1,b2,b3,b4;c1,c2,c3,c4;d1,d2,d3,d4}
and for s:
s=[C{1,1} C{2,1} C{3,2} C{4,3};C{1,3} C{2,2} C{3,3} C{4,4};...]
Tia
Tia le 16 Mai 2013
thank you

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David Sanchez
David Sanchez le 14 Mai 2013
You can try as well something like this
I=imread(your_32x32_image_name);
division = 8; % set according to yuor needs
L = size(I,1)/division;
subI = cell( L,L );
for row=1:L
for col=1:L
subI{row,col} = I( (row:row*division),(col:(col*division)) );
end
end
  1 commentaire
Tia
Tia le 16 Mai 2013
thank you

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