Replace values in matrix with NaNs?

Question

Shayma Al Ali 2023-8-24

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2012527-replace-values-in-matrix-with-nans

编辑： Voss 2023-8-24

I have two matrices with sizes 720x361. There are NaN values in matrix A that are not present in Matrix B. I'd like to insert NaNs in the same coordinates there are NaNs in matrix A. However, its not working for me.

My code:

B=rand(720,361);

[rows,cols]=find(isnan(A));

B(rows,cols)=NaN;

The "After" should have NaNs over land while keeping the values that are in the ocean. What am I doing wrong?

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Answer 1

Voss 2023-8-24

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https://ww2.mathworks.cn/matlabcentral/answers/2012527-replace-values-in-matrix-with-nans#answer_1293332

在 MATLAB Online 中打开

Do this:

B(isnan(A)) = NaN;

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Voss 2023-8-24

编辑：Voss 2023-8-24

在 MATLAB Online 中打开

Example:

% a matrix with some NaNs:
A = magic(5);
A([1 3 7 10 18]) = NaN
A = 5×5
   NaN    24     1     8    15
    23   NaN     7    14    16
   NaN     6    13   NaN    22
    10    12    19    21     3
    11   NaN    25     2     9
% another matrix the same size:
B = rand(size(A));
B_orig = B; % save the original for later
% your attempt to set elements in B to NaN, using two subscripts:
[rows,cols]=find(isnan(A))
rows = 5×1
     1
     3
     2
     5
     3
cols = 5×1
     1
     1
     2
     2
     4
B(rows,cols)=NaN
B = 5×5
       NaN       NaN    0.3915       NaN    0.9488
       NaN       NaN    0.0499       NaN    0.9620
       NaN       NaN    0.3112       NaN    0.7506
    0.0097    0.8307    0.5469    0.5113    0.2595
       NaN       NaN    0.8783       NaN    0.6373

Now B has NaNs in rows [1 2 3 5] and columns [1 2 4], which is every row in rows and every column in cols, instead of just the individual locations you wanted.

The method in my answer uses logical indexing instead:

B = B_orig; % restore original value
idx = isnan(A) % logical matrix
idx = 5×5 logical array
   1   0   0   0   0
   0   1   0   0   0
   1   0   0   1   0
   0   0   0   0   0
   0   1   0   0   0
B(idx) = NaN
B = 5×5
       NaN    0.5085    0.3915    0.8077    0.9488
    0.1175       NaN    0.0499    0.5750    0.9620
       NaN    0.2165    0.3112       NaN    0.7506
    0.0097    0.8307    0.5469    0.5113    0.2595
    0.6478       NaN    0.8783    0.1385    0.6373

You could also convert your rows and cols to linear indices and use those:

B = B_orig; % restore original value
[rows,cols]=find(isnan(A));
idx = sub2ind(size(B),rows,cols)
idx = 5×1
     1
     3
     7
    10
    18
B(idx) = NaN
B = 5×5
       NaN    0.5085    0.3915    0.8077    0.9488
    0.1175       NaN    0.0499    0.5750    0.9620
       NaN    0.2165    0.3112       NaN    0.7506
    0.0097    0.8307    0.5469    0.5113    0.2595
    0.6478       NaN    0.8783    0.1385    0.6373