[row, col] = find(isnan(YourMatrix));

The code has multiple issues that need to be fixed. Here a a couple of things to improve: The fopen documentation states that fid is an integer file identifier .... The variable fid does not contain the file data, it is merely a reference to an open file. And it is also a scalar value. Therefore your attempt to index into fid as if it were a data array doesn't make any sense. Over-complicated method of reading a text file: first textscan, then cellfun calling textscan again, all just to avoid some header lines? Instead you should read textscan 's documentation, and use the HeaderLines option, like this: fid = fopen('filename.txt', 'r'); data = textscan(fid, '%f', 'delimiter','\t', 'HeaderLines',20); fclose(fid) I also removed the CollectOutput option, as this is superfluous if there is only one format specifier. For the same reason the delimiter doesn't really make sense either. If you have multiple values per line, then you need to specify them in the formatSpec: there are plenty of examples in the documentation. Added by MathWorks Support Team : Starting in R2018b, you can use the “rmmissing” function to remove “NaN” values from an array. For example, consider the following: A = [1,NaN,2]; B = rmmissing(A) The result is the vector “B = [1 2]”. In R2018a and earlier, use the “isnan” function: A = [1,NaN,2]; B = A(~isnan(A))

Short of a mex routine that can avoid the creation of a potentially large intermediate array, I think you are stuck with isnan( ). E.g., counting the number of NaN's result = sum(isnan(MATRIX(:))); Or if you want locations, then find( ) etc. And yes, you can do similar calculations with the isinf( ) function. A mex routine to do this would not be too difficult to write, btw ...

Create a matrix of NaNs , use third output of the unique function to convert floating numbers to integer indexes and use matrix indexing to fill the matrices:

[t,~,it] = unique([t_A t_B]);
[L,~,iL] = unique([L_A L_B]);

A_new = NaN(numel(t),numel(L));
A_new(it(1:numel(t_A)),iL(1:numel(L_A))) = A;


B_new = NaN(numel(t),numel(L));
B_new(it(numel(t_A)+1:end),iL(numel(L_A)+1:end)) = B;

The efficient way to do this would be to pre-compute the size of the result, make sure that insertNanIndex was large enough to serve as a mask, and insert a into the correct indices all at once. Right now, you are literally re-allocating the entire array for every NaN. Numpy's insert function would be equally inefficient because it would be doing the same operation.

If, as in your example, the number of zeros matches the number of elements of a, you can allocate an array based on insertNanIndex and mask it directly:

result = nan(size(insertNanIndex));
result(~insertNanIndex) = a;

If the number of zeros in insertNanIndex is not equal to the size of a, you can pad or trim it, but in that case, it becomes more of a moot point as to what the whole thing means.

This is a straight forward question and given you haven't shown any attempt to solve it, I am hesitant to provide an answer ... Start by making some nonsense data n = 10; m = 5; a = randn(n, m); a(rand(numel(a), 1) < 0.25) = nan; a = 0.5377 -1.3499 0.6715 0.8884 NaN 1.8339 3.0349 -1.2075 -1.1471 NaN -2.2588 0.7254 0.7172 -1.0689 0.3192 NaN -0.0631 NaN -0.8095 0.3129 NaN 0.7147 0.4889 -2.9443 -0.8649 -1.3077 -0.2050 1.0347 1.4384 NaN -0.4336 -0.1241 NaN 0.3252 -0.1649 0.3426 NaN -0.3034 -0.7549 0.6277 3.5784 NaN 0.2939 1.3703 NaN NaN 1.4172 -0.7873 -1.7115 1.1093 The ISNAN function tells you if an element is a nan (1) or not (0). isnan(a) ans = 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 1 0 0 1 1 0 0 0 0 Now we want to sum up all the ones in each column. We have the SUM function that does that sum(isnan(a)) ans = 3 2 2 0 4 To get the final answer we want to find which column has the smallest sum. The MIN function usually returns the smallest values, but if you read the documentation, the second output argument is the index of the minimum value. [~, col] = min(sum(isnan(a))) col = 4 where the ~ notation allows for the first output argument to be ignored.

>> B1 = tb1; >> B1(A1==0) = NaN;