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Strange bug when indexing vector

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Matt Flood
Matt Flood on 14 Dec 2021
Commented: Matt Flood on 14 Dec 2021
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??
  2 Comments
Voss
Voss on 14 Dec 2021
Edited: Voss on 14 Dec 2021
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
Matt Flood on 14 Dec 2021
Thank you @Benjamin
I see it now - it's treating [3;4] as a single interable as opposed to two individual ones.

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