finding consecutive NaN in matrix

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prashant singh
prashant singh 2017-10-9
评论： Image Analyst 2018-10-18
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.

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Cedric
Cedric 2017-10-9
编辑：Cedric 2017-10-9
>> x = [1,2,3,4,NaN,NaN,NaN,7,8,9] ;
>> strfind( isnan(x), true(1,3) )
ans =
5
if you just need a "flag found", do it as follows:
>> found = ~isempty(strfind( isnan(x), true(1,3) ))
found =
logical
1
  2 个评论
Tumelo Maja
Tumelo Maja 2018-10-17
How would i apply this method to matrix? I want to find consecutive NaNs in columns of a matrix.
Image Analyst
Image Analyst 2018-10-18
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.
end

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更多回答（1 个）

Image Analyst
Image Analyst 2017-10-9
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.
  2 个评论
Cedric
Cedric 2017-10-9
To be honest, I would not have thought about using STRFIND on numbers had I not seen it first on Loren's blog back then.
Image Analyst
Image Analyst 2017-10-10
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));
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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