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sorting without moving NaNs

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How can I sort an array without moving the NaN elements?
I have A=[20 10 NaN 66 NaN 12] and I would like to get A=[66 20 NaN 12 NaN 10]. Thank you for your help. Alessio.

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Accepted Answer

Jos (10584)
Jos (10584) on 20 Dec 2013
Edited: Jos (10584) on 20 Dec 2013
Use a variant of the same trick:
A=[6 20; 10 10; 3 NaN; 20 66; 4 NaN; 7 12]
B = flipud(sortrows(A,[2 1])) ; % descending sortrows based on 2nd column
A(~isnan(A(:,2)),:) = B(~isnan(B(:,2)),:)

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Image Analyst
Image Analyst on 20 Dec 2013
Sebastiano's "Answer" moved here since it's a comment, not an answer:
Thanks a lot, it works perfectly!

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More Answers (4)

Jos (10584)
Jos (10584) on 20 Dec 2013
Edited: Jos (10584) on 20 Dec 2013
A=[20 10 NaN 66 NaN 12]
B = sort(A,'descend')
A(~isnan(A)) = B(~isnan(B))

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Azzi Abdelmalek
Azzi Abdelmalek on 20 Dec 2013
Edited: Azzi Abdelmalek on 20 Dec 2013
A=[20 10 NaN 66 NaN 12]
idx=~isnan(A);
B=sort(A,'descend');
B(isnan(B))=[];
A(idx)=B

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Jan
Jan on 20 Dec 2013
Edited: Jan on 20 Dec 2013
If A is large and contains a lot of NaN's, excluding them from the sorting can save some time:
A = [20 10 NaN 66 NaN 12]
idx = ~isnan(A);
A(idx) = sort(A(idx), 'descend');

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Jos (10584)
Jos (10584) on 20 Dec 2013
I doubt that excluding NaNs is faster, Jan. Increasing the proportion of NaNs seems to have little effect, or even a positive effect sometimes:
a = randperm(1e7) ; a(a(1:1e1)) = NaN ; tic ; sort(a) ; toc ;
a = randperm(1e7) ; a(a(1:1e5)) = NaN ; tic ; sort(a) ; toc ;

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Sebastiano delre
Sebastiano delre on 20 Dec 2013
I am sorry. Actually I wrongly state my problem. I have A=[6 20; 10 10; 3 NaN; 20 66; 4 NaN; 7 12] and I would like to get A=[20 66; 6 20; 3 NaN; 7 12; 4 NaN; 10 10]. This means I would like to sort the rows of the matrix in a descending order based on the second column and without moving the rows with the NaN elements in the second column. I would like to use sortrows. Sorry again. Alessio.

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