finding consecutive NaN in matrix

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i have a vector like [1,2,3,4,NaN,NaN,NaN,7,8,9]. I want to find if there is 3 consecutive NaN present in the vector. i can go for loop but is there more direct and easy way to do that.

Accepted Answer

Cedric Wannaz
Cedric Wannaz on 9 Oct 2017
Edited: Cedric Wannaz on 9 Oct 2017
>> x = [1,2,3,4,NaN,NaN,NaN,7,8,9] ;
>> strfind( isnan(x), true(1,3) )
ans =
if you just need a "flag found", do it as follows:
>> found = ~isempty(strfind( isnan(x), true(1,3) ))
found =
Image Analyst
Image Analyst on 18 Oct 2018
Apply it one column at a time
for k = 1 : size(m, 2)
thisColumn = m(:, k);
nanRows = isnan(thisColumn));
% Now do whatever you want to do with nanRows.

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More Answers (1)

Image Analyst
Image Analyst on 9 Oct 2017
For a more general purpose function, use isnan() and regionprops() (in the Image Processing Toolbox). It will find all groupings of Nans no matter how long or short they are. You can also specify a size range for the groups if you want, like ignore nans that are only one element long or whatever. If you want this general purpose code, let me know. Otherwise Cedric's code using the handy but little known strfind() trick is great for your given example of 3 nans in a row.
Image Analyst
Image Analyst on 10 Oct 2017
Right, I mean who would have thought that a string function could work on numbers? It's not intuitive (that's why I called it a trick).
Anyway, in case anyone is interested, here is the general case that finds lengths and indexes for all the NaN regions in the vector:
m = [1,2,3,4,NaN,NaN,NaN,7,8,9,NaN,NaN,0,NaN,9,NaN,NaN,NaN,NaN] % Sample data
nanLocations = isnan(m) % Get logical array of whether element is NaN or not.
props = regionprops(nanLocations, 'Area', 'PixelIdxList'); % Find all the regions.
% DONE! Now let's print them out
for k = 1 : length(props)
fprintf('Region #%d has length %d and starts at element %d and ends at element %d\n',...
k, props(k).Area, props(k).PixelIdxList(1), props(k).PixelIdxList(end));
Here's what shows up in the command window:
m =
1 2 3 4 NaN NaN NaN 7 8 9 NaN NaN 0 NaN 9 NaN NaN NaN NaN
nanLocations =
1×19 logical array
0 0 0 0 1 1 1 0 0 0 1 1 0 1 0 1 1 1 1
Region #1 has length 3 and starts at element 5 and ends at element 7
Region #2 has length 2 and starts at element 11 and ends at element 12
Region #3 has length 1 and starts at element 14 and ends at element 14
Region #4 has length 4 and starts at element 16 and ends at element 19

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