remove double indexvalues in loop

Hi,

I have a script where I take the maximum values out of a matrix for each column. I am using the index of the max value to get that one.

Sometimes indexmax is more then 1 value. That is the moment that i only need the highes value of A(:,e)

Now I want to avoid that i am using 2 times the same index. I want to use that index just once for the highes match.

for e = 1:length(RawDatafile)
     
     ...
     for f = 1:length(CLfile)
     ...
     for g = 1:IndRD
 ...

     percentage = (100/length2)*length3;

     A(f,e) = percentage;
     
     end
     indexmax = find(A(:,e)==max(A(:,e)));
     % this is the place where i need to add a while loop or something to avoid a double used index later in the script.
     %something like
     %if indexmax(e) == used earlier
      %   use maximum highes value of (A(indexmax(e)))
     %end

I tried the indexmax(1) but on that way its using just the first index and not the one he didn't use before.

[Bob Thompson]

Is there a specific reason for not just using max with the dimensional option?

colmaxes = max(A,[],1);

[Dion Theunissen]

If i have more indices, i need to figure out which column from the first index is the highest

[Bob Thompson]

Just to make sure I understand, you're indicating that the max values from each column may not be unique to a specific column, and you want to know the first column that contains the max?

If this is correct, then I recommend using the expanded form of unique.

[unimaxes, ia, ic] = unique(colmaxes);

'ia' will be the indices for the first appearance of each unique max value, i.e. the first column for each max value.

[Dion Theunissen]

Sorry thats not correct, maybe a bit unclear from my side.

Usually this script works, except the case when there are in column e more from the same values. In that case indexmax can have more than 1 value. Now I have to make it that way that in the 'e loop' it will not be possible to get 2 times the same index.

So in case index = 2 and the next loop index = [2, 3] it has to take 3 cause 2 is allready been.

[Bob Thompson]

Ah, ok, I think I better understand now. I recommend a check to see if the value is already within indexmax.

idm = find(A(:,e)==max(A(:,e)));    % Find index of max values
idm = idm(~ismember(idm,indexmax)); % Remove indices already recorded
indexmax(e) = idm(1);               % Record next index