Fill empty matrix with determined values

Question

James Knowles 2017-12-15

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/373050-fill-empty-matrix-with-determined-values

评论： James Knowles 2017-12-17

I am trying to enter the first local maxima of 'matrix1' into 'empty_matrix' over a certain range. There are three possible outcomes of local maxima, nothing (0), a 1X1 matrix and a 2X1 matrix. The issue is 'empty_matrix' wont fill up as desired

matrix1 = rand(50,50,500);
  matrix2 = rand(50,1);
  empty_matrix = zeros(50,50);   
  for nx=1:50
      for ny=1:50
          desired_range = squeeze(matrix1(nx,ny,1:50));
          [pks, loc] = findpeaks(desired_range(:));
          if pks == 0
              empty_matrix(ny,nx) = NaN;
          elseif pks == size(zeros(2,1))
              empty_matrix(ny,nx) = matrix2(loc(1)); %want the first local maximum
          elseif pks == size([])
              empty_matrix(ny,nx) = matrix2(loc);
          end
      end
  end

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

James Knowles 2017-12-15

Sorry I realize the example I've provided has multiple matrix size possibilities. The case I am looking for there are only the three matrix sizes listed possible.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Guillaume 2017-12-15

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/373050-fill-empty-matrix-with-determined-values#answer_296387

在 MATLAB Online 中打开

There are some very odd constructs in the code you've written:

desired_range = squeeze(...)
... = desired_range(:)

The squeeze reshapes the original matrix, the (:) reshapes it again, so the squeeze is pointless.

Two issues with:

if x == size(zeros(2,1)) %and if x = size([])

first one is you know what size(zeros(2, 1)) is, it's [2 1]. Same with (size([]), it's [0 0], so you may as well write it directly rather than constructing a fixed size matrix then getting its size.
second, x == somevector will return a logical vector the same size as somevector. So, the outcome of your test can be one of [true, true], [true, false], [false, true] or [false, false]. Most people do not know what the result of if logical_vector is, so it's almost always an error when they write that. In your case, it will happen to do the right thing, but if you wanted to test for equality of two matrices, you'd use isequal:

if isequal(pks, [2 1])

Except that's not what you want to test at all, you want to see if the number of peaks is equal to a value, in which case:

if numel(peaks) == 2

But since you're doing the same thing regardless if there is one or two peaks (take the first one), there's no point in differentiating between them. So overall, this is how I would have written your code (assuming I understood what you're trying to do:

for nx = 1:size(matrix1, 1)    %don't hardcode sizes. Query them
    for ny = 1:size(matrix, 2)   %note that traditional y is vertical and thus the 1st dimension         
        desired_range = matrix1(nx, ny, 1:50); 
        [~, loc] = findpeaks(desired_range(:));
        if isempty(loc)
            empty_matrix(ny,nx) = NaN;
        else
            empty_matrix(ny,nx) = matrix2(loc(1)); %want the first local maximum
        end
    end
end