Matrix from a for FOR loop with IF conditioning

I have the following problem:

A = [1 2 3; 4 5 6; 7 8 9]

for i = 1:n

      if rem(i,2)== 0
         x = fliplr(D(i,:))
      else x = D(i,:)
      end

   B(i) = x

end

I'm trying to get the B matrix that should look like

B = [1 2 3; 6 5 4; 7 8 9]

but I get the following error message instead

In an assignment  A(I) = B, the number of elements in B and I must be the same.

Error in back_and_forth (line 16)
   B(i) = x

Could anyone help me with this?

Answer 1

[Azzi Abdelmalek]

A = [1 2 3; 4 5 6; 7 8 9]
A(2:2:end,:)=fliplr(A(2:2:end,:))

%Or if you want to do it with a for loop

A = [1 2 3; 4 5 6; 7 8 9]
n=size(A,1)
B=A;
for i = 1:n
  if rem(i,2)== 0
      B(i,:) = fliplr(A(i,:))
  end
end

Answer 2

[Jos (10584)]

n = size(A,1)
...
B(i,:) = x
...

[Jos (10584)]

or using indexing alone

B = A ; % keep the original
B(2:2:end,:) = B(2:2:end,end:-1:1)

Answer 3

[Geoff Hayes]

Hi Chris,

I'm guessing that n should be 3 in this example, and that in the code D is set (at some point) with A.

So the problem is with your assignment, B(i)=x. You are assigning a single element of B to a row vector of three elements. This may be fine if B were defined to be a cell array (and you had written B{i}), but since you want it to be a matrix, then your assignment should mimic how you got x but in the opposite "direction".

Note that the assignment for x is x = D(i,:) or x = fliplr(D(i,:)). In either case you extract all columns (using the :) from the ith row of D. So when you assign this vector to B it should be the "same" - set all columns (using the :) of the ith row of B to x:

B(i,:) = x;

Note that to avoid the B is growing at each iteration warning, you could just initialize B at the beginning to:

B = zeros(size(D));

Hope that this helps!

Geoff