How do I find the number of occurrences of NaN and the corresponding subscripts in an array

1 Ansicht (letzte 30 Tage)
A = NaN 100 101 102 103 104
201 2 7 3 2 2
202 NaN 8 4 5 6
203 NaN NaN 2 3 5
205 3 4 2 6 4
I have a matrix with the first row and first column being headers. I would like to know the no. occurences and subscripts on NaN.
To find no. of occurences, I did
number_of_nan = sum(sum(isnan(A(2:end,2:end))))
Also,
logical_array = isnan(A(2:end,2:end));
numel(logical_array(logical_array == 1));
Is there a simpler/better way. Also how do I find the subscripts of the NaN elements in the array ?
  2 Kommentare
José-Luis
José-Luis am 9 Jan. 2013
What do you mean by subscripts? The linear indexes or the row and column position? The indices relative to what? To the entire array or to the array without column and headers?
Ms. Mat
Ms. Mat am 9 Jan. 2013
Bearbeitet: Ms. Mat am 9 Jan. 2013
I would like to know the row and column position. For the given example matrix A,
202, 100
203, 100
203, 101
or
3, 2
4, 2
4, 3
Thank You

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José-Luis
José-Luis am 9 Jan. 2013
Bearbeitet: José-Luis am 9 Jan. 2013
Another option:
logical_array = A(2:end,2:end) ~= A(2:end,2:end)
num_NaN = sum(logical_array(:));
idx = find(logical_array); %I am not sure this is what you want, please see my comment to your question
EDIT so that was not what your wanted after all. For that:
To get row and column position, according to the header:
logical_array = A(2:end,2:end) ~= A(2:end,2:end)
idx = find(logical_array);
[dim(1) dim(2)] = size(A);
[ii,jj] = ind2sub(dim-1,idx);
your_position = [A(ii+1,1) A(1,jj+1)'];

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