Deleting Rows inbetween NaN values.

I got a row vector that looks like this: V = [ 1 4 5 NaN 0 9 3 1 3 NaN 1 5 3 0 7 NaN 7 4 2 1 NaN 9 4 ]. I want to delete all the columns in between the NaN values, and place the remaining columns in a cell array, it will look like this: Output = { [ 1 4 5 ] [ 1 5 3 0 7 ] [ 9 4 ] }. Not all the vectors in the array will be the same length, so a simple reshape will not work. There are thousands of lines as well. Thank You.

[Peng Li]

It wasn't quite clear that the [1 5 3 0 7] subset is also in between two NaN values, while you keep it?

[Chad]

Yes, I want every other subset between the NaN values to be kept, starting with the first one, as the cell array Output suggest.

Answer 1

[Stephen23]

>> V = [1,4,5,NaN,0,9,3,1,3,NaN,1,5,3,0,7,NaN,7,4,2,1,NaN,9,4];
>> X = isnan(V);
>> Y = cumsum(X);
>> F = @(n) V(~X&(n==Y));
>> C = arrayfun(F, 0:2:max(Y), 'UniformOutput',false);
>> C{:}
ans =
   1   4   5
ans =
   1   5   3   0   7
ans =
   9   4

Answer 2

[Vinai Datta Thatiparthi]

Hi Chad,

Understanding the problem from the single example that you mentioned, this following approach was one of the simplest that I could think of -

v = [ 1 4 5 NaN 0 9 3 1 3 NaN 1 5 3 0 7 NaN 7 4 2 1 NaN 9 4 ]; % Input array
idN = isnan(v); % For NaNs in array
idF = find(idN == 1); % Find indices of NaNs

out = {}; % Output cell array

for i = 1:2:numel(idF)
    if i == 1
        out{end+1} = v(1:idF(1)-1); % For first index
    else
        out{end+1} = v(idF(i-1)+1:idF(i)-1); % For the rest
    end
end

if rem(numel(idF),2) == 0
    out{end+1} = v(idF(end)+1:end); % At the end
end

Hope this helps!