How to cut an array and reshape it while keeping the content in order?
11 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
Joseph Schmidt
el 5 de Jul. de 2016
Respondida: Jos (10584)
el 6 de Jul. de 2016
I have an array similar to what's below. How do I cut and move it so that after the 11th column (were it begins to go to 1 again) so that it's a 10x11 matrix instead of 5x22?
The reshape function doesn't work as it keeps the 1st column, pushes the 2nd column down and below the 1st column, keeps the 3rd column, and pushes the 4th column below the 2nd column. It messes up the data. Basically I want to cut the latter half of the data from column 12:end and place it below the 1:11 columns.
(5x22) T1 =
1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 0 0 0
1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0
1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 0
1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 0
(10x11)T1 =
1 1 1 1 1 1 1 1 0 0 0
1 1 1 1 1 1 1 0 0 0 0
1 1 1 1 1 1 1 1 1 1 0
1 1 1 1 1 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 0
1 1 1 1 1 1 1 1 0 0 0
1 1 1 1 1 1 1 0 0 0 0
1 1 1 1 1 1 1 1 1 1 0
1 1 1 1 1 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 0
0 comentarios
Respuesta aceptada
Star Strider
el 5 de Jul. de 2016
This is how I would do it:
T1 = [T1(:,1:11); T1(:,12:end)]
2 comentarios
Star Strider
el 5 de Jul. de 2016
My pleasure.
The easiest way is to use the size function to define the column length.
This works with your ‘T1’ matrix, and should work for the others:
T1 = [T1(:,1:fix(size(T1,2)/2)); T1(:,fix(size(T1,2)/2)+1:end)]
The fix call may not be absolutely necessary, but I always include it to prevent problems.
Más respuestas (3)
Stephen23
el 6 de Jul. de 2016
Editada: Stephen23
el 6 de Jul. de 2016
This code can easily be changed to split the matrix into two, three, or more blocks. It assumes that the number of columns is divisible by this number of blocks.
>> blk = 2;
>> tmp = reshape(T1,size(T1,1),[],blk);
>> reshape(permute(tmp,[1,3,2]),[],size(tmp,2))
ans =
1 1 1 1 1 1 1 1 0 0 0
1 1 1 1 1 1 1 0 0 0 0
1 1 1 1 1 1 1 1 1 1 0
1 1 1 1 1 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 0
1 1 1 1 1 1 1 1 0 0 0
1 1 1 1 1 1 1 0 0 0 0
1 1 1 1 1 1 1 1 1 1 0
1 1 1 1 1 0 0 0 0 0 0
1 1 1 1 1 1 1 1 1 1 0
Example
We first define a function:
function out = moveblock(mat,blk)
tmp = reshape(mat,size(mat,1),[],blk);
out = reshape(permute(tmp,[1,3,2]),[],size(tmp,2));
end
and test this on a matrix:
>> T = ones(3,1)*(1:12)
T =
1 2 3 4 5 6 7 8 9 10 11 12
1 2 3 4 5 6 7 8 9 10 11 12
1 2 3 4 5 6 7 8 9 10 11 12
>> moveblock(T,2)
ans =
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
7 8 9 10 11 12
7 8 9 10 11 12
7 8 9 10 11 12
>> moveblock(T,3)
ans =
1 2 3 4
1 2 3 4
1 2 3 4
5 6 7 8
5 6 7 8
5 6 7 8
9 10 11 12
9 10 11 12
9 10 11 12
>> moveblock(T,4)
ans =
1 2 3
1 2 3
1 2 3
4 5 6
4 5 6
4 5 6
7 8 9
7 8 9
7 8 9
10 11 12
10 11 12
10 11 12
0 comentarios
Jos (10584)
el 6 de Jul. de 2016
This is easy using a cell array as an intermediate step
T = repmat(1:12,3,1) % example array
S = size(T)
n = 3
if mod(S(2),n)>0
C = mat2cell(T, S(1), repmat(S(2)/n,1,n))
Tout = cat(1,C{:})
else
disp('Not possible.')
Tout = []
end
0 comentarios
Ver también
Categorías
Más información sobre Loops and Conditional Statements en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!