Matrix and find

Question

Salvatore Turino 2012-1-15

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https://ww2.mathworks.cn/matlabcentral/answers/26167-matrix-and-find

Hello actually i have a random matrix done of a lot of 0 and some 1. i need to save the position (i,j) of the first 1 for each row/column. If i have a matrix like this one

   1     0     0     0     1     0     0
   0     1     0     0     0     0     0
   0     1     0     0     0     0     0

doing a(i)=find(matrix(i,:),1) i got a=[2,3,3], and this is ok.

Instead if i have this matrix:

   0     0     0     0     0     0     0
   0     0     0     0     0     0     0
   0     1     0     0     0     0     0

doing the same thing with the find i got error.

Improper assignment with rectangular empty matrix.

why?

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Image Analyst 2012-1-15

What do you want or expect to get when you have no 1's in the row? Maybe -1 or something? It's got to be something.

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Answer 1

Jan 2012-1-15

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You have to decide, what you want as output, if a row does not contain any 1.

n = size(matrix, 1);
a = zeros(1, n);   % Pre-allocate!
for i = 1:n
  found = find(matrix(i,:), 1);
  if isempty(found)
    a(i) = NaN;  % Or Inf, or -1, or 0, or whatever?!
  else
    a(i) = found;
  end
end

[EDITED]: You can define the default value in the pre-allocation to make the code 2 lines shorter:

n = size(matrix, 1);
a = nan(1, n);   % Pre-allocate!
for i = 1:n
  found = find(matrix(i,:), 1);
  if ~isempty(found)
    a(i) = found;
  end
end

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Salvatore Turino 2012-1-16

thank you :) this is was i was looking for. yesterday evening i did it in a different way but your code is more compact and functional. thank you

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Answer 2

the cyclist 2012-1-15

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For the first matrix, when i=1, "find(matrix(1,:),1)" looks at the first row of the matrix, and looks for the first column that has a non-zero in it. That's column 2, so you are doing a(1)=2, which is fine.

For the second matrix, when i=1, the same command returns the empty matrix, because there is no non-zero in the first row. Therefore, you are trying to do a(1)=[], which gives the error you see.

Edit in response to your comment:

Here's a different way from Jan's, in which only the rows that actually have non-zeros are displayed:

x = [0 1 0 0 0 1 0 0 ...
     0 0 1 0 0 0 0 0 ...
     0 0 1 0 0 0 0 0];     
[m,n]=find(x);
[rowsWithNonZero,indexToFindColumn] = unique(m,'first');
columnsOfFirstNonZero = n(indexToFindColumn);
disp('row:')
disp(rowsWithNonZero)
disp('column:')
disp(columnsOfFirstNonZero)