Borrar filtros
Borrar filtros

Split big matrix in many submatrices having same size

3 visualizaciones (últimos 30 días)
Mehdi Kooli
Mehdi Kooli el 4 de Nov. de 2022
Comentada: Jilin Zhang el 21 de Dic. de 2022
Hello,
I have a big matrix 396*2600 and I want to split it to many matrices 30*120 as follow : (let's consider this exemple on a small matrix A ans split it to 3*2 matrices):
A = [1 9 17
2 10 18
3 11 19
4 12 20
5 13 21
6 14 22
7 15 23
8 16 24];
results wanted: A1 = [1 9
2 10
3 11]
A2 = [9 17
10 18
11 19]
A3 = [2 10
3 11
4 12]
A3 = [10 18
11 19
12 20] and so on ...
Remarks: I need a solution without for loop, I'm looking for a Matlab command giving this results.
Thanks
  6 comentarios
Mehdi Kooli
Mehdi Kooli el 23 de Nov. de 2022
Movida: Matt J el 23 de Nov. de 2022
Hello Andrei Bobrov, I saw you answered similar question (https://fr.mathworks.com/matlabcentral/answers/357482-reshaping-a-2d-matrix-into-a-3d-matrix-layers-of-square-matrices-with-specific-order-no-loop?s_tid=srchtitle)
Can you please take a look here , maybe you have the solution.
Mehdi Kooli
Mehdi Kooli el 9 de Dic. de 2022
Movida: Matt J el 9 de Dic. de 2022
Hello @Andrei Bobrov maybe you have the answer ?

Iniciar sesión para comentar.

Respuestas (2)

John D'Errico
John D'Errico el 4 de Nov. de 2022
Editada: John D'Errico el 4 de Nov. de 2022
So many times this gets asked for. DON'T DO IT. Instead, learn to use arrays, of many types. For example, just make it into a 3 dimensional array, where each plane of that array is one of the desired sub-arrays. That requires relatively little more than understanding how to index arrays.
A = [1 9 17
2 10 18
3 11 19
4 12 20
5 13 21
6 14 22
7 15 23
8 16 24];
Here you want to generate all 3x2 contiguous subarrays. The size of A is
[r,c] = size(A)
r = 8
c = 3
So there will be 6*2 such 3x2 sub-arrays to generate.
Where are the elements of A stored in memory? In what order? Understanding this, and how tools like sub2ind work in MATLAB allows you to build these arrays easily.
ind1 = (1:3)' + [0,r];
ind2 = (1:r-2) + r*[0:c-2]';
B = reshape(A(ind1(:)-1 + ind2(:)'),3,2,[]);
Now it is simple to acces any of those subarrays as we have created, and do so programmatically. The array B is of size:
size(B)
ans = 1×3
3 2 12
There are 12 such sub-arrays.
B(:,:,1)
ans = 3×2
1 9 2 10 3 11
B(:,:,2)
ans = 3×2
9 17 10 18 11 19
B(:,:,12)
ans = 3×2
14 22 15 23 16 24
  5 comentarios
Mehdi Kooli
Mehdi Kooli el 12 de Dic. de 2022
@Stephen23 Thanks a lot, it resolves the problem for this example but I need to adapt the code for a big matrix A= 426 x 2904 using a sliding window 60 x 240
I don't understand the value 2 in the code (see comments below)
Here is the code :
rect_l=3; % rows lenght of my sliding window
rect_c=2; % columns lenght of my sliding window
% Matrix example 8 x 3
A = [1 9 17
2 10 18
3 11 19
4 12 20
5 13 21
6 14 22
7 15 23
8 16 24];
[r,c] = size(A);
x1 = (1:rect_l)';
y1 = [0,r];
ind1 = bsxfun(@plus,x1,y1);
x2 = (1:r-2); % Why the value 2 ?
y2 = r*[0:c-2]'; % Why the value 2 ?
ind2 = bsxfun(@plus,x2,y2);
ind3 = bsxfun(@minus,ind1(:),1);
ind4 = ind2(:)';
ind5 = bsxfun(@plus,ind3,ind4);
B = reshape(A(ind5),rect_l,rect_c,[]);
Can you please tell me how to modify the code for the new Matrix size and sliding window.
Jilin Zhang
Jilin Zhang el 21 de Dic. de 2022
A=randi([0 10],426,2904)
[r1, c1]=size(A);
r2=60 %target row #
c2=240 %target column #
i=1
j=1
k=1
while i*r2<r1
while j*c2<c1
B(k,:,:)=A((i-1)*r2+1:i*r2,(j-1)*c2+1:j*c2);
j=1+j
k=1+k
end
i=1+i
end

Iniciar sesión para comentar.


Matt J
Matt J el 7 de Nov. de 2022
Editada: Matt J el 7 de Nov. de 2022
What I'm trying to do is to split the matrix to submatrices and then for each submatrix, calculate the median value and store the result in another matrix.
The problem simplifies greatly if you choose a submatrix size with odd dimensions, in which case you can use medfilt2. Otherwise, ordfilt2 would allow you the increased flexibility of using any dimensions.

Categorías

Más información sobre Logical en Help Center y File Exchange.

Etiquetas

Productos


Versión

R2015b

Community Treasure Hunt

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

Start Hunting!

Translated by