Problem with matrix manipulation

Assume user input data as below. I define my matrix is cost. The matrix i created is 3 by 3 matrix. So the matrix should form like this:

cost = [c11 c12 c13
   c21 c22 c23
   c31 c32 c33]

Since I want to display set of row, I do it like this:

c1 = cost(1,:); % it will become c1 = c11 c12 c13
c2 = cost(2,:); % it will become c2 = c21 c22 c23
c3 = cost(3,:); % it will become c3 = c31 c32 c33

Then I want the value in the matrix. I do it like this.

c11 = cost(1,1);
c12 = cost(1,2);
c13 = cost(1,3);
c21 = cost(2,1);
c22 = cost(2,2);
c23 = cost(2,3);
c31 = cost(3,1);
c32 = cost(3,2);
c33 = cost(3,3);

So this is the equation that I want to use for this type of matrix.

lambda = ((8*c13*c23*c33*Pdt)+(4*c12*c23*c33)+(4*c13*c22*c33)+(4*c13*c23*c32)) ./ (4*c23*c33)+(4*c13*c33)+(4*c13*c23));

So my problem is, if I want to make 4 by 3 matrix, and it would generate a matrix like this:

cost = [c11 c12 c13
  c21 c22 c23
  c31 c32 c33
  c41 c42 c43]

The equation that I want to use for this matrix(4 by 3) is quite different. How can I do that? Do I need to use if else statement? Or do while? Can anyone help me solve this? Can anyone create the code?

Answer 1

[Rob Graessle]

You can use the SIZE function to check how many rows are in the cost matrix, then use IF/ELSE to change the equation used for lambda.

if size(cost, 1) == 3
    % Use lambda equation for 3 rows      
    lambda = ((8*cost(1,3)*cost(2,3)*cost(3,3)*Pdt)+ (4*cost(1,2)*cost(2,3)*cost(3,3)+(4*cost(1,3)*cost(2,2)*cost(3,3)+(4*cost(1,3)*cost(2,3)*cost(3,2)) ./ (4*cost(2,3)*cost(3,3)+(4*cost(1,3)*cost(3,3)+(4*cost(1,3)*cost(2,3));
elseif size(cost, 1) == 4
    % Use lambda equation for 4 rows 
    lambda = ...;
end

Also note that you don't need to create new variables c11, c12, etc. - it's not very memory-efficient.

Answer 2

[Walter Roberson]

It is not recommended that you use this approach at all. Please see the FAQ for alternatives.

Answer 3

[slumberk]

@rob

what is the meaning of size(cost,1) == 3?? Does it have 3 row n column??

@walter

thx for the info =)

[Rob Graessle]

The '1' argument means you want the size along the first dimension (rows). So this line checks that cost has 3 rows. If you wanted to check the number of columns you would use size(cost,2). Or to get both:

[r,c] = size(cost)

where r is the number of rows and c is columns.

[slumberk]

@rob

thx rob =)