Matlab: find neighboring instances of NaN without loops - matlab

Question

Essentially, I have a matrix of data with many "holes" represented by NaN, and I want to retrieve the indices of all NaN's that are clustered fewer than 4 times in a single column.
e.g. with the matrix:
A =
23 12 NaN 56 60 21 NaN
60 56 94 22 45 NaN NaN
23 55 19 83 NaN NaN NaN
NaN NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN NaN
84 99 43 32 89 12 NaN
76 92 73 47 22 12 10
23 55 12 93 61 94 20
NaN NaN NaN NaN NaN NaN NaN
41 16 83 39 82 37 43
14 78 92 40 81 29 60
it would return:
ans =
[4; 5; 6; 10; 16; 17; 18; 22; 25; 28; 29; 30; 34; 40; 41; 42; 46; 58; 70; 82]
So far, I have a vector with the indices of all the NaN values from
nan_list=find(isnan(A(:)))
but I don't know how to extract sequential numbers from that vector without using loops, which would be too expensive. I also tried something similar to the answer posted by b3 here, by switching all NaN's to a value that doesn't appear in the matrix, but that code was not as transferable for other data sets.
Thanks for any suggestions!

Answer 1

Code
N = 4; %// Fewer than clusters of N or N+ NaNs are to be detecteed
nan_pos = isnan(A) %// Find NaN positions as a binary array
conv_res = conv2(double(nan_pos),[0 ones(1,N)]')==N %//' Perform convolution
start_ind = find(conv_res(N+1:end,:)) %// Find positions where clusters of N or N+ NaNs start
nan_pos(unique(bsxfun(#plus,start_ind,[0:N-1])))=0 %// Get positions of all those clustered N or N+ NaNs and set them in NaN position array as zeros
out = find(nan_pos) %// Finally the desired output
Example
As an example, let's try this code on a slightly different input that would hopefully test out various aspects of the problem -
A = [
23 12 NaN 56 60 21 NaN
60 56 94 22 45 NaN NaN
23 55 19 83 NaN NaN NaN
NaN NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN NaN
84 99 43 32 89 12 NaN
76 92 73 47 22 12 10
23 55 12 93 61 94 20
NaN NaN NaN NaN NaN NaN NaN
41 NaN NaN 39 82 37 43
14 78 NaN 40 81 NaN 60]
Now, let's assume that we are looking to find indices of cluster fewer than 3 NaNs. Thus editing N as 3 in the code, the output is -
out =
10 22 23 25 46 58 70 72 82
This makes sense when we look into the input.

Answer 2

This should work:
[rows, ~] = size(A);
maxNansPerCol = 4;
% find which columns have few enough NaNs
Anans = isnan(A);
nansInCols = sum(Anans);
qualifyingCols = nansInCols <= maxNansPerCol;
% zero the other columns
mask = repmat(qualifyingCols,rows,1);
B = Anans .* mask;
% get the NaN locations
indices = find(B(:));
(Apologies if something is slightly off--I don't have MATLAB on this computer to test it)