Convert numeric 2D array to array of orders of values

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.)

Answer 1

[Dyuman Joshi]

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

Additionally -

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

[dormant]

Many thanks. I was trying to avoid the loop approach, but you've done it much better than I would have.

[Dyuman Joshi]

You are welcome!

My initial thought was to look for a vectorized solution as well, but I couldn't come up with one.