Remove NaN inside a loop cycle

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luca
luca on 16 Sep 2019
Answered: Star Strider on 16 Sep 2019
Given the following code
V = [ 19 15 11 2 16 3 1 18 14 3 18 19 1 13 16 14 19 15 4 14 3 1 2 16 4 3 1 19 3 20 4 13 1 15 2 18 4 1 19 17 3 1 13 3 4 17 18 19 14 ;
19 16 11 2 16 15 1 18 14 3 18 19 14 2 16 14 19 15 4 14 3 11 2 16 4 3 1 19 3 20 4 13 1 4 2 18 4 1 19 17 3 19 13 3 4 17 18 19 14 ];
A= [0 15 11 2 4 0 3 1 13;
14 0 16 0 0 0 0 0 0;
0 0 0 0 8 0 0 0 0;
0 0 18 0 0 0 0 0 0;
0 0 19 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0];
%V = [V(1,:), NaN(1, mod(-size(V,2), 10))];
for i= 1:size(V,1)
VVV = [V(i,:), NaN(1, mod(-size(V,2), 10))];
%assert(mod(numel(V), 10) == 0, 'V length is not a multiple of 10');
V10 = reshape(VVV, 10, []); %rearrange V in batches of 10
[~, where] = ismember(V10, A); %find location of elements in A
[~, col] = ind2sub(size(A), where); %convert location into column index
[~, order] = sort(col, 1); %and sort by column order
X (i,:) = reshape(V10(order + (0:size(order, 1):numel(order)-1)), 1, []); %and rearrange columns of V10 according to that order
B(i,:) = X(~isnan(X(i,:)))';
end
I'm generating a matrix X where inside are contained some NaN. How can I remove the NaN inside the rows?
the code
B(i,:) = X(~isnan(X(i,:)))';
does not work!
The number of NaN is the same in each rows, so we don't need to use cells
May someone help me?

Accepted Answer

Star Strider
Star Strider on 16 Sep 2019
Try this:
B(i,:) = X(i,~isnan(X(i,:)));

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