Reshape matrix by taking two consecutive rows every two rows
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Hello everyone,
I have seen quite some questions here on the forum that are about matrix reshaping. However, for my particular problem I did not find an answer. So, here we are:
I have got a matrix that looks like
[1 1 1 1;
2 2 2 2;
3 3 3 3;
4 4 4 4;
5 5 5 5;
6 6 6 6;
7 7 7 7;
8 8 8 8]
Is it somehow possible to reshape it to
[1 1 1 1 3 3 3 3;
2 2 2 2 4 4 4 4;
5 5 5 5 7 7 7 7;
6 6 6 6 8 8 8 8]
with the built-in functions like reshape, permutate, flip,...?
There would always be the way with for loops but I thought it might be neater (and possibly faster) to do it with the built-in functions.
Thanks for your help.
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采纳的回答
Adam Danz
2021-3-23
编辑:Adam Danz
2021-4-7
x must be an n*m matrix where n is divisible of 4.
M is the reformatted matrix with size k*j where k=n/2 and j=m*2.
x = [1 1 1 1;
2 2 2 2;
3 3 3 3;
4 4 4 4;
5 5 5 5;
6 6 6 6;
7 7 7 7;
8 8 8 8];
rowIdx = [1 3 2 4]' + [0:4:size(x,1)-1];
M = reshape(x(rowIdx(:),:)', size(x,2)*2, [])'
Another example,
x = [1 1 1 1;
2 2 2 2;
3 3 3 3;
4 4 4 4;
5 5 5 5;
6 6 6 6;
7 7 7 7;
8 8 8 8
9 9 9 9
10 10 10 10
11 11 11 11
12 12 12 12];
rowIdx = [1 3 2 4]' + [0:4:size(x,1)-1];
M = reshape(x(rowIdx(:),:)', size(x,2)*2, [])'
To get from M back to x,
k = reshape(M',size(M,2)/2,[]);
rowIdx = [1 3 2 4]' + (0:4:size(k,2)-1);
xOrig = k(:,rowIdx(:))';
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更多回答(1 个)
Taimoor Hasan Khan
2021-3-23
A = [1 1 1 1;
2 2 2 2;
3 3 3 3;
4 4 4 4;
5 5 5 5;
6 6 6 6;
7 7 7 7;
8 8 8 8];
A = cat( 2, [ A(1:2, :); A(5:6, :) ], [ A(3:4, :); A(7:8, :) ] )
% vertical concat. #1 % vertical concat #2
Not sure how you could generalize the process without loops though, sorry...
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