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

1 回表示 (過去 30 日間)
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 件のコメント
José-Luis
José-Luis 2013 年 1 月 9 日
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 2013 年 1 月 9 日
編集済み: Ms. Mat 2013 年 1 月 9 日
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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採用された回答

José-Luis
José-Luis 2013 年 1 月 9 日
編集済み: José-Luis 2013 年 1 月 9 日
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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