Strange bug when indexing vector

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Matt Flood 2021-12-14

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https://ww2.mathworks.cn/matlabcentral/answers/1610790-strange-bug-when-indexing-vector

评论： Matt Flood 2021-12-14

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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??

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Voss 2021-12-14

编辑：Voss 2021-12-14

在 MATLAB Online 中打开

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 2021-12-14