Sorting elements of a matrix ignoring NaN

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Stephen
Stephen 2016-3-21
编辑： Azzi Abdelmalek 2016-3-21
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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Azzi Abdelmalek
Azzi Abdelmalek 2016-3-21
编辑：Azzi Abdelmalek 2016-3-21
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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