# Deleting elements of an array with consecutively occurring NaN values

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okoth ochola on 18 May 2023
Commented: Dyuman Joshi on 19 May 2023
Hello, may inqure about something small. Am trying to filter a huge data I obatined from my measurements. Iwould like to delete array elements with NaN value and that occur consecutively in an array based on a thresh hold value. For example, i have matrix A, i can determine the position of the non-empty (not NaN) cells,as shown in the code below. I would then go ahead and find the diffenrece between the elemnts of NotNaN matrix,if the diffenerence is equal or greater than 3, then all the elements in matrix A, in between the positions defined by the NotNaN matrix (the adjacents elements whose diffenrence is greater or equal to 3) are deleted. So that the final output would be A= [34 45 10 22 36 33 28 21 NaN 20 98 NaN]
A = [43 NaN NaN NaN 45 NaN NaN 10 22 NaN NaN 36 33 28 21 NaN 20 98 NaN];
NotNaN=find(~isnan(A))
diff(NotNaN)
NotNaN =
1 5 8 9 12 13 14 15 17 18
ans =
4 3 1 3 1 1 1 2 1

Dyuman Joshi on 18 May 2023
Note that this removes any NaN groups occuring at the start or at the end of the data.
%Modified data
A = [43 NaN NaN NaN 45 NaN NaN 10 22 NaN NaN 36 33 28 21 NaN 20 98 NaN 17 NaN NaN NaN NaN NaN ...
2 3 NaN 5 7 NaN 13 NaN NaN];
idx = find(isnan(A))
idx = 1×18
2 3 4 6 7 10 11 16 19 21 22 23 24 25 28 31 33 34
%Indices corresponding to single NaNs
out = setdiff(setdiff(idx,idx+1),idx-1)
out = 1×4
16 19 28 31
%Remove consecutive NaNs
A(setdiff(idx,out)) = []
A = 1×20
43 45 10 22 36 33 28 21 NaN 20 98 NaN 17 2 3 NaN 5 7 NaN 13
Dyuman Joshi on 18 May 2023
I am unable to think of a vectorized solution to this approach yet, I will update if I find one. Meanwhile, here is a loop approach that achieves what you want to do, but it will be slow for large inputs
%Modified data
A = [NaN NaN NaN NaN 43 NaN NaN NaN 45 NaN NaN 10 22 NaN NaN 36 33 28 21 NaN 20 98 NaN 17 NaN NaN NaN NaN NaN ...
2 3 NaN 5 7 NaN 13 NaN NaN NaN NaN];
%Adding a number to the end
A = [A 0];
disp(A)
Columns 1 through 33 NaN NaN NaN NaN 43 NaN NaN NaN 45 NaN NaN 10 22 NaN NaN 36 33 28 21 NaN 20 98 NaN 17 NaN NaN NaN NaN NaN 2 3 NaN 5 Columns 34 through 41 7 NaN 13 NaN NaN NaN NaN 0
%First NaN
idx = find(isnan(A),1);
%First number after NaN
k = find(~isnan(A(idx+1:end)),1)+idx;
thresh = 3;
while ~isempty(k)
if k-idx>=thresh
A(idx:k-1)=[];
k=idx;
end
idx = find(isnan(A(k+1:end)),1)+k;
k = find(~isnan(A(idx+1:end)),1)+idx;
end
%Remove the number
A = A(1:end-1);
disp(A)
43 45 NaN NaN 10 22 NaN NaN 36 33 28 21 NaN 20 98 NaN 17 2 3 NaN 5 7 NaN 13
okoth ochola on 19 May 2023
thank you @Dyuman Joshi

### More Answers (1)

Stephen23 on 19 May 2023
A = [NaN,NaN,NaN,NaN,43,NaN,NaN,NaN,45,NaN,NaN,10,22,NaN,NaN,NaN,NaN,36,33,28,21,NaN,20,98,NaN,17,NaN,NaN,NaN,NaN,NaN,2,3,NaN,5,7,NaN,13,NaN,NaN,NaN,NaN]
A = 1×42
NaN NaN NaN NaN 43 NaN NaN NaN 45 NaN NaN 10 22 NaN NaN NaN NaN 36 33 28 21 NaN 20 98 NaN 17 NaN NaN NaN NaN
N = 3;
X = isnan(A(:));
D = cumsum([1;diff(X)~=0]);
F = @(v)sum(v)>N;
Y = grouptransform(ones(size(D)),D,F);
A(X&Y) = []
A = 1×25
43 NaN NaN NaN 45 NaN NaN 10 22 36 33 28 21 NaN 20 98 NaN 17 2 3 NaN 5 7 NaN 13
##### 2 CommentsShow NoneHide None
okoth ochola on 19 May 2023
Edited: okoth ochola on 19 May 2023
Wow, thank you @Stephen23; However the my matlab is not recognizing the function grouptransform(), I guess it was added in more recent versions of matlab. Am using 2018a
Dyuman Joshi on 19 May 2023
Yes, grouptransform was added in just the next version of MATLAB - 2018b.

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