Sorting elements of a matrix ignoring NaN
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Dear All,
I have the following 4x4 matrix have containing numbers and NaN:
have = [5 NaN 4 9;
4 0 NaN 9;
-6 NaN 2 3;
1 7 NaN -3]
I would like to assign a rank (ascending/descending) to the elements of each colum of the matrix have. In the sorting procedure, the NaNs should be ignored and elements with the same value (column 4) should have the same (tied) rank.That is, I would like to obtain the following matrices:
want_ascend = [4 NaN 2 3;
3 1 NaN 3;
1 NaN 1 2;
2 2 NaN 1]
want_descend = [1 NaN 1 1;
2 2 NaN 1;
4 NaN 2 2;
3 1 NaN 3]
My attempt involving the sort function does not successfully ignore the NaN (which are ranked too).
[~,attempt]=sort(have,1,'descend')
attempt =
1 1 2 1
2 3 4 2
4 4 1 3
3 2 3 4
Any help would be highly appreciated. Thanks!
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Accepted Answer
Azzi Abdelmalek
on 21 Mar 2016
Edited: Azzi Abdelmalek
on 21 Mar 2016
have = [5 NaN 4 9;
4 0 NaN 9;
-6 NaN 2 3;
1 7 NaN -3]
idx=isnan(have)
[n,m]=size(have)
[have_ascend,have_descend]=deal(zeros(n,m))
for k=1:m
v=have(:,k)
[~,~,kk]=unique(v)
have_ascend(:,k)=kk;
end
have_ascend(idx)=nan
You can get have_descend from have_ascend
have_descend=bsxfun(@minus,max(have_ascend)+1,have_ascend)
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