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

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Daria Ivanchenko
Daria Ivanchenko 2020 年 10 月 23 日
編集済み: Peter Perkins 2020 年 11 月 20 日
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!

回答 (2 件)

Saurav Chaudhary
Saurav Chaudhary 2020 年 10 月 27 日
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;
%Otherwise we can move ahead
else
R = R + 1;
end
end

Peter Perkins
Peter Perkins 2020 年 11 月 20 日
編集済み: Peter Perkins 2020 年 11 月 20 日
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)

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