# Find the certain amount of values between NaNs and make another variable from these values

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Daria Ivanchenko on 23 Oct 2020
Edited: Peter Perkins on 20 Nov 2020
Hi!
I have a matrix A which has 50 rows and 15000 columns with some numbers and sometimes NaNs between them. In each of these rows I want to find periods of time when the continuous length of non-NaN values was not shorter then 270 (but it can 300, 500 or longer, whatever). I want to take 270 first values out from each of these epochs and make another variable from it. Is it somehow possible?
Just an example:
A = 1 2 3 NaN NaN NaN NaN NaN 4 5 NaN NaN NaN 6 7 8 9 10 NaN NaN
NaN NaN 1 2 3 4 NaN NaN NaN NaN 5 NaN NaN NaN 6 7 8 9 10 NaN
1 NaN NaN NaN 2 3 4 NaN NaN 5 6 NaN NaN 7 8 9 10 11 NaN NaN
I want to find the epochs when the number of values between NaN is 3 or more (for example). And then I take the first 3 elements of the epoch and put it to the new line of the new variable. In the end my new variable should look like this:
result = 1 2 3
6 7 8
1 2 3
6 7 8
2 3 4
7 8 9
I hope I explained clearly.
Thanks for any help!

Saurav Chaudhary on 27 Oct 2020
Yes, it is possible. You can go through the code given below to get an idea, but again modify it according to your application:
Input matrix:
A = [ 1 2 3 NaN NaN NaN NaN NaN 4 NaN NaN NaN NaN 6 7 8 9 10 NaN NaN
NaN NaN 1 2 3 4 NaN NaN NaN NaN 5 NaN NaN NaN 6 7 8 9 10 NaN
1 NaN NaN NaN 2 3 4 5 NaN 5 6 NaN NaN 7 8 9 10 11 3 4];
Converting the matrices into 1-D vector, so as to traverse it easily:
[r,c] = size(A);
n = r * c;
reshape(A,[1,n]);
A= A';
Assigning left(L) and right(R) handles:
L= 1 ;
R= 2 ;
Specify limit you want between two NaN's:
Lim= 3 ;
Main body of function:
while R <= n
%it takes care of the case when there are two consecutive number but NaN is next
if isnan( A( R ) )
L = R;
R = R + 1;
end
%when there are consecutive numbers between NaN's
if R- L + 1 == Lim
Tem = zeros(1, Lim);
i = 1;
while i <= Lim
Tem(i) = A(L);
L = L + 1;
i = i + 1;
end
while R<=n && ~isnan( A( R ) )
R = R + 1;
end
L = R;
R = R + 1;
disp(Tem)
%When either of the L or R is NaN we can skip them
elseif isnan( A( L ) ) || isnan( A( R ) )
L = R;
R = R + 1;
else
R = R + 1;
end
end

Peter Perkins on 20 Nov 2020
Edited: Peter Perkins on 20 Nov 2020
There's no need to loop over the 750K elements. Do this on each row x of your data, i.e. loop from 1:50.
Try stepping through this code in the debugger with these test rows
x = [1 2 3 NaN NaN NaN NaN NaN 4 5 NaN NaN NaN 6 7 8 9 10 NaN NaN]
%x = [1 2 3 NaN NaN NaN NaN NaN 4 5 NaN NaN NaN 6 7 8 9 10 11 12]
%x = [NaN NaN 1 2 3 4 NaN NaN NaN NaN 5 NaN NaN NaN 6 7 8 9 10 NaN]
%x = [NaN NaN 1 2 3 4 NaN NaN NaN NaN 5 NaN NaN NaN 6 7 8 9 10 11]
and you will have learned a useful MATLAB idiom.
startsStops = find(diff(isnan(x)))+1; % <= the key
if ~isnan(x(1))
startsStops = [1 startsStops];
end
if ~isnan(x(end))
startsStops = [startsStops length(x)+1];
end
starts = startsStops(1:2:end);
stops = startsStops(2:2:end);
runLengths = stops - starts;
minLength = 3;
longRuns = (runLengths >= minLength);
starts = starts(longRuns);
runLengths = runLengths(longRuns);
numLongRuns = length(starts);
runs = starts + (0:minLength-1)';
xruns = reshape(x(runs),length(starts),minLength)