Understanding :error To RESHAPE the number of elements must not change while manipulating matrix

Hi, i need some explaination, i run this code and the input matrix 40*11 and resulted 41*11 matrix. I try to put it into a new matrix by using reshape and has error.

function [] = cacl_mean(G)
k = 4
j = 1
i=1
[numRows,numCols] = size(G) %numRows =40,numCols =11

    while j <= k
        for n = i:40 %iteration dr data row 1 hingga 40
            if (G(i,11)) == j %check data col ke 11 ada value j=1@2@3@4
                L=G(i,:)
                 i=i+1
            else
                i=i+1
            end
        end
        n=1
        i=1
        disp(n)
        disp(j)
        j=j+1
    end 
    kk = reshape(L,41,11); % error To RESHAPE the number of elements must not change.
end

Answer 1

[Guillaume]

It's very unclear what your code is meant to do, it makes no sense.

Anyway, in your loop your creating (sometimes!) L as a vector consisting of a row of your G matrix. If the condition is true several times in your loop, this L will contain only the last row for which it is true. Also note that if the condition is never true, then L won't exist and you'll have an error telling you it's undefined later on.

So, if L exists, it's a vector corresponding to one row of G. According to your comments G has 11 columns, so L is 1x11. Obviously, it can't be reshaped into a 41x11 matrix.

You should learn to debug your programs. By stepping through your code one line at a time, you'd have seen the problem immediately.

Notes:

• You query the size of the input matrix, a good thing to do, but then completely ignore that and hardcode it in the loop. So in the end your code only works for a 40x11 matrix.
• Your function doesn't return anything
• it's unclear why you're using a while loop, when for j = 1:k would be clearer.
• It's unclear why the i=i+1 is in both branches of the if. May as well put it afterward and write the line only once.
• More importantly it's very unclear what you meant to do with your for n=i:40, which here is exactly equivalent to for n = 1:40 since the value of i is always 1 when the loop starts. Additionally you never use n. Your loops are exactly equivalent to:

for j = 1:4
    for i = 1:40  %would be better as for i = 1:numRows
        if G(i, 11) == j
            L = G(i, :);
        end
    end
end

Again, this does the same as what you wrote, only it's a lot clearer. Of course, it's unlikely to be what you meant to do.