Help with matrix indexing.

C is a matrix with i rows and 5 columns. i=1:16

I am interested in finding the value of C(i-1,:).

Using the following matlab code, I try to direct matrix c row i-1 any column to c row 16 any column when i equals 1.

if i==1
  c(i-1,:)=c(16,:);
end

 Error: Subscript indices must either be real positive integers or logicals.

Problem: It gives me above error message as it seems always recognize the c(i-1,:)=c(0,:) first and therefore ignore the statement after that which is equalling to the c(16,:).

Any idea how to solve it?

Answer 1

[the cyclist]

You seem to be treating this in a sorta symbolic way trying to tell MATLAB that the c(i-1) element is really element c(16,0) when i==1. However, what you have done in practice is an assignment statement,

>>  c(0,:) = c(16,:)

which is not valid because you can't assign to element c(0,:).

I think the easiest way to accomplish what you want is to use the CIRCSHIFT command:

>> index = (1:16)'
>> down_one_index = circshift(index,1)

The guts of CIRCSHIFT, in your case, is the following use of MOD:

>> mod((0:15)-1, 16)+1

This creates the periodic indexing that you want. Hope that helps.

the cyclist

Answer 2

[Siddharth Shankar]

if "i" is 1, then you can not index into c with the row index "i-1". MATLAB uses 1-based indexing. You need to "special case" your code to avoid indexing into row 0, or a row value that is negative, as the error message suggests.

Answer 3

[Lucy]

Hi, I solved the problem by using

if i==1

cellindex=16;

elseif i~=1

cellindex=i-1;

c(cellindex,:)

end

Thank you both anyway.

[the cyclist]

Glad you found a solution. Be aware that applying logic functions to handle special cases is in general going to be slower than vectorized solutions like the one I proposed.

[Lucy]

I can see your point now. Thanks

Answer 4

[Paulo Silva]

Lucy there's no need for any if statements or other fancy functions

nrows=16 %choose the number of rows (horizontal lines)
indrow=2:nrows+1 %create the index of the rows, the trick is here (+1)
C(indrow-1,:) %no problem now, indrow-1 is never zero

[the cyclist]

Lucy's trying tell MATLAB that the row "below" row 1 should be row 16. (Periodic boundary condition.) I don't see how your indexing does this. "indrow-1" seems to just be 1:16. Do I misunderstand?

[Paulo Silva]

Now I understand, thanks for the good explanation :) forget about my code, it just rotates the rows 180º