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Logical indexing a 2D array into a 2D array

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Eric Smith
Eric Smith el 19 de Abr. de 2018
Respondida: Wouter Verstraelen el 19 de Abr. de 2021

I am having trouble finding a way to remove all rows and columns that are all zeros from a 2D matrix to get a resulting 2D matrix using logical indexing.

Z = [56,0,0,0,0,55;
      0,0,0,0,0,0;
     27,0,0,0,0,0;
      0,0,0,0,0,0;
      0,0,0,0,0,0;
     100,0,0,0,0,25];
zidx = Z ~= 0;
Z2 = Z(zidx);

results in column vector of:

Z2 = [56;27;100;55;25]

However I really need it to be:

Z2 = [56,55; 27,0; 100,25];

Any help is greatly appreciated.

Respuesta aceptada

Jan
Jan el 19 de Abr. de 2018
Editada: Jan el 19 de Abr. de 2018

Easier and faster without a loop:

Z = [56,0,0,0,0,55;
      0,0,0,0,0,0;
     27,0,0,0,0,0;
      0,0,0,0,0,0;
      0,0,0,0,0,0;
    100,0,0,0,0,25];
Z = Z(:, any(Z, 1));  % Keep only columns with any non-zero value
Z = Z(any(Z, 2), :);  % Keep only rows with any non-zero value

If you really want Z = [56,27,100; 55,0,25] as output, append:

Z = Z.'
  2 comentarios
Eric Smith
Eric Smith el 19 de Abr. de 2018
Thanks. This works and is closer to what I was thinking as a solution.
Guillaume
Guillaume el 19 de Abr. de 2018
is closer to what I was thinking as a solution
This is the best solution!

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Más respuestas (2)

Wouter Verstraelen
Wouter Verstraelen el 19 de Abr. de 2021
there's actually a very short to do this that hasn't been mentioned so far
Z =Z.*(Z~= 0)

Bob Thompson
Bob Thompson el 19 de Abr. de 2018

So, to clarify, you want to have your columns turned into rows?

Because you are looking to roughly keep the row and column locations I would suggest using a for loop to keep the non-zero values from each row. Before that though, because you have different numbers of values you're going to need to either initialize the output array as the max size in zeroes, or to pad as you add new rows.

Z2 = [];
for k = 1:size(Z,1); % Easier to do row wise because of needing to pad rows, rather than columns.
 row = Z(k,Z(k,:)~=0);
 if isempty(row);  
 elseif isempty(Z2);
  Z2 = row;
 elseif size(Z2,2)<length(row); % Check if previous row was smaller than new row
  Z2(:,size(Z2,2)+1:length(row)) = 0;
  Z2 = vertcat(Z2,row);
 elseif size(Z2,2)>length(row); % Check if previous row was bigger than new row
  row(length(row)+1:size(Z2,2)) = 0;
  Z2 = vertcat(Z2,row);
 else
  Z2 = vertcat(Z2,row);
 end
end
Z2 = Z2';
  1 comentario
Eric Smith
Eric Smith el 19 de Abr. de 2018
Editada: Eric Smith el 19 de Abr. de 2018
This works, and sorry that I mistyped column major for output matrix... output should be: Z = [56,27,100; 55,0,25];

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