How to swap specific elements position of a matrix

Hello everyone. I have a matrix like A = [ 1 0 0 2; 0 3 4 0; 5 6 0 0; 0 0 7 8 ]

I want a new matrix like B = [ 5 0 0 8; 0 6 7 0; 1 3 0 0; 0 0 4 2 ]

Is it possible to do that? Thank you.

Answer 1

[DGM]

The problem is a bit underdefined. I'm going to make the following assumptions:

• There are always exactly two nonzero elements per column
• The goal is to merely swap the nonzero elements in each column regardless of sorting

A = [ 1 0 0 2; 0 3 4 0; 5 6 0 0; 0 0 7 8 ]

% assuming there are exactly two nonzero elements per column
nz = sort(A,1);
B = zeros(size(A));
B(A~=0) = flipud(nz(end-1:end,:))

gives

A =
     1     0     0     2
     0     3     4     0
     5     6     0     0
     0     0     7     8
B =
     5     0     0     8
     0     6     7     0
     1     3     0     0
     0     0     4     2

[Jenjen Ahmad Zaeni]

It works, thank you very much. Yes, sorry for unclear explanation, the goal is to swap the nonzero elements on every column and keep the zero elements on their position.

[Jenjen Ahmad Zaeni]

Hello DGM, are you still can answer it? Sorry but i make mistakes. Should i make a new thread or reply here? Well, maybe i write here first. I want to make a matrix like this:

A =  
-225.2565         0         0  226.1974 -225.2565  225.2565         0
-225.2565         0  225.2565  226.1974         0         0 -225.2565
        0 -257.1251  248.9089         0         0  248.9089 -248.9089
  20.8466         0         0  -29.3149   20.8466         0   20.8466
        0 -219.0711         0  218.4656         0  218.4656 -218.4656

to become like this:

B =  
  20.8466         0         0  218.4656   20.8466  218.4656         0
-225.2565         0  248.9089  -29.3149         0         0 -218.4656
        0 -219.0711  225.2565         0         0  248.9089   20.8466
-225.2565         0         0  226.1974 -225.2565         0 -248.9089
        0 -257.1251         0  226.1974         0  225.2565 -225.2565

note that i can't use something like idx here, cause i generate matrices randomly everytime i running the script, so the position of zeros will be different. The goal is to swap the bottom element to top, top element to bottom, and so on, like the illustration above, and keep the zeros remain on their places. Thank you very much for your help.

[DGM]

If the number of nonzero elements varies per column, it gets a bit more complicated. This might be able to be simplified, but I just used a loop and logical indexing.

A = [-225.2565 0 0 226.1974 -225.2565 225.2565 0;
    -225.2565 0 225.2565 226.1974 0 0 -225.2565;
    0 -257.1251 248.9089 0 0 248.9089 -248.9089;
    20.8466 0 0 -29.3149 20.8466 0 20.8466;
    0 -219.0711 0 218.4656 0 218.4656 -218.4656];

B = zeros(size(A));
for n = 1:size(A,2)
    m = A(:,n)~=0;
    B(m,n) = flipud(A(m,n));
end
B

B = 5×7

20.8466 0 0 218.4656 20.8466 218.4656 0 -225.2565 0 248.9089 -29.3149 0 0 -218.4656 0 -219.0711 225.2565 0 0 248.9089 20.8466 -225.2565 0 0 226.1974 -225.2565 0 -248.9089 0 -257.1251 0 226.1974 0 225.2565 -225.2565

[Jenjen Ahmad Zaeni]

I've tried this with other random value and it works perfectly like this. Once again, thank you very much. Have a great day.

>> A

A =

         0   25.5540 -227.4337  -25.5540  227.4337    1.4323         0
         0 -257.3220         0         0         0 -269.4361  227.4337
   52.9148  272.7692         0         0 -257.3220   -1.4323  257.3220
         0         0 -233.1677   -1.4323         0         0         0
  257.6266  -32.0834  269.4361  -25.5540 -269.4361         0         0

>> B

B =

         0  -32.0834  269.4361  -25.5540 -269.4361   -1.4323         0
         0  272.7692         0         0         0 -269.4361  257.3220
  257.6266 -257.3220         0         0 -257.3220    1.4323  227.4337
         0         0 -233.1677   -1.4323         0         0         0
   52.9148   25.5540 -227.4337  -25.5540  227.4337         0         0