How do I assign unique values to a i*j matrix/grid?

I have a grid of i rows and j columns. For example, 3 rows and 5 columns. However, the origin is in the bottom left corner, not the top left corner as for matrices.

I have an equation, b= i + j, which I want to use to assign a unique value to each location on the matrix, not to replace the individual values of the matrix/grid. (the matrix doesn't have values, it is just a rectangle divided into a grid). e.g the bottom left grid square will have the value 2 assigned to it.

I have tried using this code:

    m=[1 2 3 4 5]
    n=[3 2 1]

    for i=n(1):n(length(n))
        for j= m(1):m(length(m))
              l(i,j)= i + j
        end
    end

I know this doesn't work, because l hasn't been assigned. I'm not sure how to do this however. I want to be able to type l(1,1) for example, and to be able to see which value has been assigned to this point.

I hope I've expressed this clearly enough. I appreciated any help.

Thank you

Answer 1

[Geoff Hayes]

Sarah - the problem is in the outer for loop

 for i=n(1):n(length(n))

Since initial value (initval) is n(1)=3, end value (endval) is n(length(n))=1, and index is i, the loop becomes

 for i=3:1

which from for description states that increments the index variable from initval to endval by 1, and repeats execution of program statements until index is greater than endval.

Since the index immediately greater than the end value (3>1) we don't enter the for loop. I think what you want to do is something like the following

 % pre-allocate memory to the output matrix
 output = zeros(length(n),length(m));

 for u=n
     for v=m
         output(u,v) = u + v;
     end
 end

The above allows us to iterate with the indices u and v (I switched these from your i and j since MATLAB uses these two to (also) represent the imaginary number) over each element in n and m respectively. Basically, the statement for u=n allows us to let u be each element in n. The output from the above is

 output =

     2     3     4     5     6
     3     4     5     6     7
     4     5     6     7     8

which is not quite what you wanted since you stated that the the origin is in the bottom left corner, not the top left corner as for matrices.

So we can do this transformation as follows

 numRows = length(n);
 numCols = length(m);

 % pre-allocate memory to the output matrix
 output = zeros(numRows,numCols);

 for u=n
     for v=m
         output(numRows-u+1,numCols-v+1) = u + v;
     end
 end

The output from the above is

 output =

     8     7     6     5     4
     7     6     5     4     3
     6     5     4     3     2

Note that we are assuming that the range in each vector, m and n, is from 1 to the length of the vector.

[Sarah]

Thank you! Not exactly what I was looking for but it gave me a few ideas. I really appreciate the response.

Answer 2

[Azzi Abdelmalek]

    m=[1 2 3 4 5]
    n=[3 2 1]
    kn=numel(n);
    km=numel(m);
    l=zeros(kn,km); % preallocate
    for i=1:numel(n)
        for j= 1:numel(m)
              l(i,j)= n(i) +m(j);
        end
    end

%or without for loop

    m=[1 2 3 4 5]
    n=[3 2 1]
    [x,y]=meshgrid(m,n)
    l=x+y

[Sarah]

Thank you very much! I appreciate the response.