rearrangement rows of matrices
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Suppose:
A = repmat((1:15)',1,5);
Now I want to rearrange A such that it becomes matrix of size 5 by 15 in following form:
B = [A(1:5,:), A(6:10,:), A(11:15,:)];
The problem is that size of matrix A in my work is too big and its size may change in each iteration. So I can't use a fix command like above to evaluate B. On the other hand, using a for loop is too much time consuming. I tried function "reshape". But, it does not work well.
Is there any function to do this? or how should I call function reshape to get what I want?
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採用された回答
Bruno Luong
2020 年 11 月 17 日
A = reshape(A,[5 3 5]);
A = permute(A,[1 3 2]);
A = reshape(A,[5 15]);
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その他の回答 (3 件)
Rik
2020 年 11 月 17 日
The easiest way is probably to use a cell as an intermediary step:
A = repmat((1:15)',1,5);
B = [A(1:5,:), A(6:10,:), A(11:15,:)];
d=5;%or is this size(A,2)?
C=mat2cell(A,...
d*ones(1,size(A,1)/d),...
size(A,2));
C=C.';
C=cell2mat(C);
isequal(B,C)
This would all be much simpler if your example array is actually this repetitive, as you could use repelem.
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KSSV
2020 年 11 月 17 日
A = repmat((1:15)',1,5);
B = [A(1:5,:), A(6:10,:), A(11:15,:)];
[r,c] = size(A) ;
n = 3 ;
C = permute(reshape(A',[c,r/n,n]),[2,1,3]);
D = reshape(C,5,[]) ;
0 件のコメント
Andrei Bobrov
2020 年 11 月 17 日
A = reshape(1:75,15,[]);
a = 5;
m = size(A,1);
B = reshape(permute(reshape(A,a,m/a,[]),[1,3,2]),a,[]);
2 件のコメント
Antonio Carvalho
2022 年 2 月 2 日
@Andrei Bobrov how are you. Can i reshape two long vectors? one has a lenght completely different to another and from a spefic point (time) compare them using a plot figure?
Rik
2022 年 2 月 2 日
You can't. If you don't have as many x-values as you have y-values it is not possible to match them up and create a plot. You need to make sure you can make pairs of your x-y-values.
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