Strange bug when indexing vector

The following piece of code works as I want it to.

It sorts the elements in a 2x2 matrix and gives equal rank order to elements with the same value.

x = [2.3333 2.3333; 2.0000 2.3333];
[y, z] = sort(x(:))

y = 4×1

2.0000 2.3333 2.3333 2.3333

z = 4×1

2 1 3 4

if any(diff(y)==0)
    for n = find(diff(y')==0)+1
        z(n) = z(n-1);
    end
end
disp(z)

     2
     1
     1
     1

So x is sorted as [2, 2.33, 2.33, 2.33] and the elements are given the ranks [2, 1, 1, 1].

However, if I don't transpose y in the for loop, the code gives the wrong output, i.e., the rank is [2, 1, 1, 3].

x = [2.3333 2.3333; 2.0000 2.3333];
[y, z] = sort(x(:))

y = 4×1

2.0000 2.3333 2.3333 2.3333

z = 4×1

2 1 3 4

if any(diff(y)==0)
    for n = find(diff(y)==0)+1
        z(n) = z(n-1);
    end
end
disp(z)

     2
     1
     1
     3

As far as I can tell, there is no logical reason why not transposing y vector should produce a different output.

Can someone please explain why this is happening??

[Voss]

for loops in MATLAB loop over the columns of the variable you tell it. So, for example, this:

for i = [1 2 3]
    display(i);
end

i = 1

i = 2

i = 3

is as expected (columns of a row vector are scalars), but the following only loops once and i is the whole column vector:

for i = [1; 2; 3]
    display(i);
end

i = 3×1

1 2 3

Generalizing to a 2d matrix:

x = magic(3);
for i = x
    display(i);
end

i = 3×1

8 3 4

i = 3×1

1 5 9

i = 3×1

6 7 2

So in your second example (without transposing y), n takes the value [3; 4] and the loop iterates once, rather than the loop iterating twice with n taking 3 and then 4, which is what it does in the first example.

[Matt Flood]

Thank you @Benjamin

I see it now - it's treating [3;4] as a single interable as opposed to two individual ones.