converting a matrix into a column vector using only while-end loop

Hi all, I'm trying to convert a matrix into a column vector. I also need to do this only using while loops. Please help. Here is my code:

function [A] = func4(M)
[m,n] = size(M);
A = zeros(m*n,1);

i = 1;
j = 1;
c = m*n;
while i <= m
    cx = c;
    A(cx,1) = M(m, n);
    while j <= n
        A(j,1) = M(i, j)
        j = j + 1;
        
    end
    
    break
    
end
M = [1 2 3; 4 5 6]
B = func4(M)

[Jon]

First of all you can do the whole thing with just one statement

M = A(:)

Even if you were going to do it with loops it would be better to use a for loop than a while as you already know how many iterations you need to do.

Is this for a homework problem that requires you to use while loops?

[Hrvoje Sarinic]

Yes. I'm doing it for my college. They want us to make a function which, when executed, does the same thing you described (M = A(:)).

Answer 1

[Jan]

This sounds like a homework question and you have shown, what you have tried so far.

function A = func4(M)
[m,n] = size(M);
A     = zeros(m*n, 1);

The start is fine. An alternative would be to obtain the number of elements directly by numel(M), but size() is okay.

% [EDITED] Former code collected the elements in rowwise order,
%          now it is columnwise: i and j are interchanged.
k = 1;      % additional counter for the output
c = m*n;
i = j;
while j <= n
   i = 1;   % Reset j counter inside the loop
   while i <= m
        A(k) = M(i, j);  % k counter for A, i & j counter for M
        i = i + 1;
        k = k + 1;
   end
   j = j + 1;
end
end

The break is not needed. The only problem was the k counter for the output and the missing increasing of the i counter.

[Hrvoje Sarinic]

Thank you very much. Yes it is a homework assignment, and we were not allowed to use any other integrated functions except while, size and zeros so i couldn't put numel(M). Thank you for that tip anyway!

[Hrvoje Sarinic]

A(k) = M(i, j); % k counter for A, i & j counter for M

Also one small question about this part of the code: When i put A(k) without the second index, does matlab automatically assume it is 1? What I mean is will these 2 be equal then: A(k) = A(k,1) ?

Thank you in advance.

[Jan]

Simply try it. This works even in the Matlab interpreter hiere in the forum.

If only 1 index is provided, Matlab uses it as "linear index". This is the index as if the complete array would be a vector. See:

A = [11, 12; 13, 14]

A = 2×2

11 12 13 14

A(1)

ans = 11

A(2)

ans = 13

A(3)

ans = 12

A(4)

ans = 14

[Jon]

One other point. If you truly want to emulate A = M(:), then you should note that this operation turn the matrix into a column, "columnwise", meaning it stacks the columns of M. So for your example you would get A (1) = 1 A(2) = 4, A(3) = 2, A(4) = 5 etc.

I think the code above goes "rowwise", so it stacks the rows of M, and you get A (1) = 1 A(2) = 2,... A(4) = 4 etc

[Jan]

@Jon: You are right. I've interchanged i and j now to fix this.