Returning Incorrect index of array

hi,

The simple code below try to find the min value and its index, excluding 0

x           =   [0 0.5 4 5];
[r,index]   =   min(x(x>0))

The min value is correctly founded as 0.5. However, why the index returns as 1, it suppose to return 2

r =

    0.5000
index =

     1

Please help to solve. TQVM

[Stephen23]

Without changing any data values:

x = [0,0.5,4,5];
idx = find(x);
[val,idy] = min(x(idx));
val

val = 0.5000

idx(idy)

ans = 2

[Khairul Nur]

tq stephen for the code, really appreciate that!

Answer 1

[Bjorn Gustavsson]

It returns 1 for the index because the way you call min, x(x>0) becomes the array [0.5 4 5], and the first element of that array is the smallest.

HTH

[Khairul Nur]

understood, there is any way still excluding the zero but to get the index as 2?

[Scott MacKenzie]

I'm not sure if there's a one-liner, but this will work:

x = [0 0.5 4 5];
x(x==0) = realmax;
[~, idx] = min(x)

idx = 2

[Khairul Nur]

tq scott ur the best!

Answer 2

[Viranch Patel]

So if you write y = x(x>0), then it will take value only which is greater than 0 from the x and put it into y.

x = [0 0.5 4 5];
y = x(x>0);
[r,index] = min(x(x>0));

So y will be equal to [0.5000 4.0000 5.0000].

Now if you find minimum then the index will be 1 not 2.

I hope you get the answer:)

[Khairul Nur]

understood, there is any way still excluding the zero but to get the index as 2?

[Viranch Patel]

If I'm guessing correct, you want to get the index of the lowest positive value. Then you can do this using a single for loop in which you can have two variables, one will store the minimum value and one will store the index.

index = 0;

x = [0 0.5 4 5];
min_value = max(x);
len = length(x);

for i = 1:len
    if x(i) < min_value && x(i) > 0
        min_value = x(i);
        index = i;
    end
end

This is quite straight-forward. So you can edit this according to your need.

[Khairul Nur]

tq viranch.. great! its working