correction in order of elements in matrix obtained from reshape array

Question

Karanvir singh Sohal 2021-4-3

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/791489-correction-in-order-of-elements-in-matrix-obtained-from-reshape-array

评论： Karanvir singh Sohal 2021-4-3

采纳的回答： Matt J

在 MATLAB Online 中打开

Hello everyone!

below is the code for calculation of C

A=[4 4 4; 8 8 8]
B=[16 12 8]
[mA,nA] = size(A);
[mB,nB] = size(B);
index=0;
for i=1:mA
  for j=1:nB
      index=index+1;
   c(index)=(min(A(i,j),B(1,j))/max(A(i,j),B(1,j)));
  end
end
C=[reshape(c,[],nB)]

I'm obtaining this matrix

C =  [0.2500   0.5000   0.6667
      0.3333   0.5000   1.0000]

but i want results as

C =  [0.2500   0.3333   0.5000
      0.5000   0.6667   1.0000]

2 个评论
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David Fletcher 2021-4-3

It's due to the way reshape fills the reshaped matrix from the elements of the original. To get it to do what you want you could reshape to a 3x2 matrix and then transpose

reshape(c,[],numel(c)/nB)'

ans =

0.2500 0.3333 0.5000

0.5000 0.6667 1.0000

Karanvir singh Sohal 2021-4-3

Thanks @David Fletcher

This works fine. :)

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Answer 1

Matt J 2021-4-3

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/791489-correction-in-order-of-elements-in-matrix-obtained-from-reshape-array#answer_665909

在 MATLAB Online 中打开

A loop-method,

A=[4 4 4; 8 8 8];
B=[16 12 8];
c=min(A,B(1,:))./max(A,B(1,:))
c = 2×3
    0.2500    0.3333    0.5000
    0.5000    0.6667    1.0000

1 个评论
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Karanvir singh Sohal 2021-4-3

This is great!!!!

No need of for loop.

I gonna check the efficiency of this solution for different size of the matrix before using. :)

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Answer 2

Matt J 2021-4-3

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/791489-correction-in-order-of-elements-in-matrix-obtained-from-reshape-array#answer_665904

在 MATLAB Online 中打开

A=[4 4 4; 8 8 8]
A = 2×3
     4     4     4
     8     8     8
B=[16 12 8]
B = 1×3
    16    12     8
[mA,nA] = size(A);
[mB,nB] = size(B);
index=0;
for j=1:nB
for i=1:mA
      index=index+1;
   c(index)=(min(A(i,j),B(1,j))/max(A(i,j),B(1,j)));
  end
end
C=reshape(c,[],nB)
C = 2×3
    0.2500    0.3333    0.5000
    0.5000    0.6667    1.0000