# Convert numeric 2D array to array of orders of values

1 vue (au cours des 30 derniers jours)
dormant le 25 Jan 2023
Commenté : Dyuman Joshi le 31 Jan 2023
I have an array of numbers, like this:
positions =
14 25 65 20 16 15 17 16
14 26 46 0 12 0 14 5
0 0 46 13 11 11 11 17
14 25 49 11 15 17 10 11
0 0 19 15 16 20 11 13
18 4 48 20 12 12 12 24
How can I create a similar-sized array with the numbers changed to their order in each column, ie
orders =
1 2 6 4 5 3 6 4
1 3 2 NaN 2 NaN 5 1
NaN NaN 2 2 1 1 2 5
1 2 5 1 4 4 1 2
NaN NaN 1 3 5 5 2 3
4 1 4 4 2 2 4 6
This needs to cope with values that are equal, giving them both the same place and then skipping a place for the next value.
(The orders array above was created by hand, so might have errors.)
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### Réponse acceptée

Dyuman Joshi le 27 Jan 2023
Modifié(e) : Dyuman Joshi le 31 Jan 2023
I couldn't think of a vectorized solution (yet), so here is the for loop approach.
If and when I find a vectorized solution, I will update my solution.
pos = [14 25 65 20 16 15 17 16
14 26 46 0 12 0 14 5
0 0 46 13 11 11 11 17
14 25 49 11 15 17 10 11
0 0 19 15 16 20 11 13
18 4 48 20 12 12 12 24];
for idx=1:size(pos,2)
%for operating on non zeros values
non=pos(:,idx)~=0;
%indices of unique values with reference to the column
[~,~,c]=unique(pos(non,idx));
for i=2:max(c)
%counting repetitions of each indices, ct - count
ct=nnz(c==(i-1));
if ct>1
%modifying indices accordingly
c(c>i)=c(c>i)+1;
c(c==i)=ct+i-1;
end
end
pos(non,idx)=c;
end
%assign Nan to zeros
pos(pos==0)=NaN
pos = 6×8
1 2 6 4 5 3 6 4 1 4 2 NaN 2 NaN 5 1 NaN NaN 2 2 1 1 2 5 1 2 5 1 4 4 1 2 NaN NaN 1 3 5 5 2 3 4 1 4 4 2 2 4 6
Since unique() identifies each Nan as a different element, updating zeros to NaN before would not have been helpful in this approach
unique([1 -1 NaN 0 -1 -1 0 NaN])
ans = 1×5
-1 0 1 NaN NaN
(Ideally) You should use tolerance based method to check for any values in MATLAB, rather than direct equality.
y=0;
abs(y-0)<1e-1
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dormant le 31 Jan 2023
Many thanks. I was trying to avoid the loop approach, but you've done it much better than I would have.
Dyuman Joshi le 31 Jan 2023
You are welcome!
My initial thought was to look for a vectorized solution as well, but I couldn't come up with one.

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