CODEWITHSUNDEEP
arraysmatlabnan
user758114
user758114

Reputation: 354

Removing NaN Values from Array and shift left

I have a 34 x 1096 array that has NaN values.

 A = 
 NaN    0.2500     NaN    0.3750       NaN
 NaN    0.1100     NaN    0.4310     0.1250
 NaN    0.1250    0.2500  0.3750     0.4310

And I want 

 A = 
0.2500   0.3750       NaN     NaN       NaN
0.1100   0.4310     0.1250    NaN       NaN
0.1250   0.2500     0.3750   0.4310     NaN

Whats a simple way to do this?

Upvotes: 4

Views: 282

Answers (3)

Divakar
Divakar

Reputation: 221574

Code

 A = [
 NaN    0.2500     NaN    0.3750       NaN
 NaN    0.1100     NaN    0.4310     0.1250
 NaN    0.1250    0.2500  0.3750     0.4310]

[M,N] = size(A)
[~,col1] = sort(~isnan(A),2,'descend')

row1 = repmat(1:M,N,1)'; %%//'
restructured_indices = sub2ind(size(A),row1(:),col1(:))
A = reshape(A(restructured_indices),M,N)

Output

A =
    0.2500    0.3750       NaN       NaN       NaN
    0.1100    0.4310    0.1250       NaN       NaN
    0.1250    0.2500    0.3750    0.4310       NaN

Upvotes: 1

Luis Mendo
Luis Mendo

Reputation: 112689

[~, jj] = sort(isnan(A), 2);
B = A(bsxfun(@plus, (1:size(A,1)).', (jj-1)*size(A,1)));

Upvotes: 2

bla
bla

Reputation: 26069

a simple way will be to use a for loop with ~isnan on each row, for example:

B=NaN(size(A));
for n=1:size(A,1)
    B(n,1:sum(~isnan(A(n,:))))=A(n,~isnan(A(n,:)));
end    

B =
0.2500    0.3750       NaN       NaN       NaN
0.1100    0.4310    0.1250       NaN       NaN
0.1250    0.2500    0.3750    0.4310       NaN

you can then assign A=B if you must... and yes this can be done without a for loop, but why bother in this case?

Upvotes: 1

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