Convert individual values in a matrix to letters and keep remaining as numbers

Hi,

I'm trying to fill in characters corresponding to elements in the array airline_names into the array Usage using the below code:

 Usage = num2cell(CounterUsage);
airline_names = cellstr(['AC_ '; 'FI_ '; 'QR_ '; 'SK_C'; 'SQ_ '; 'STB '; 'STE '; 'TG_ ']);

 for i=1:33
    for j=1:288
        if isequal(Usage(i,j),0) ;

        else
            Usage(i,j)=num2cell(double('AB'));
        %else
            %Usage(i,j) = num2str(airline_names(CounterUsage(i,j)));
        end
    end
end

Usage is a matrix of values from 0 to 8 where I want all non zero elements to be replaced by the corresponding element in airline_names, e.g. Usage(i,j) = 1 then 1 should be replaced by airline_names(1) = 'AC_ '

However, the above code doesn't work due to what I think might be some type/format errors..

Please help!:D

Best Regards Maria and Christian

Answer 1

[Azzi Abdelmalek]

airline_names = cellstr(['AC_ '; 'FI_ '; 'QR_ '; 'SK_C'; 'SQ_ '; 'STB '; 'STE '; 'TG_ '])
usage=randi(8,8,1)
out=airline_names(usage)

[Christian]

In principle what I want to do, except that the usage matrix should contain both these airline_names and zeros on all other positions. This was why is tried looping through all elements in the usage matrix in other to check all of them a replace the values if they where not equal to zero. And furthermore I wanted to replace them with the corresponding element in the airline_names matrix.

So what I need is a matrix in which I can look up values (and a way to do so) and have both numbers and characters.

[Azzi Abdelmalek]

airline_names = cellstr(['AC_ '; 'FI_ '; 'QR_ '; 'SK_C'; 'SQ_ '; 'STB '; 'STE '; 'TG_ '])
usage=randi(8,288,33)-1
idx=find(usage==0)
idx1=setdiff(1:numel(usage),idx)
out=cell(size(usage))
out(idx)=num2cell(0)
out(idx1)=airline_names(usage(idx1))

[Christian]

Thanks man, that did the job!!!