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Logical indexing a 2D array into a 2D array

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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.

Risposta accettata

Jan
Jan il 19 Apr 2018
Modificato: Jan il 19 Apr 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 Commenti
Eric Smith
Eric Smith il 19 Apr 2018
Thanks. This works and is closer to what I was thinking as a solution.
Guillaume
Guillaume il 19 Apr 2018
is closer to what I was thinking as a solution
This is the best solution!

Accedi per commentare.

Più risposte (2)

Wouter Verstraelen
Wouter Verstraelen il 19 Apr 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 il 19 Apr 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 Commento
Eric Smith
Eric Smith il 19 Apr 2018
Modificato: Eric Smith il 19 Apr 2018
This works, and sorry that I mistyped column major for output matrix... output should be: Z = [56,27,100; 55,0,25];

Accedi per commentare.

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