Returning Incorrect index of array
4 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
Khairul Nur
il 9 Lug 2021
Commentato: Khairul Nur
il 22 Lug 2021
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
2 Commenti
Stephen23
il 12 Lug 2021
Without changing any data values:
x = [0,0.5,4,5];
idx = find(x);
[val,idy] = min(x(idx));
val
idx(idy)
Risposta accettata
Bjorn Gustavsson
il 9 Lug 2021
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
3 Commenti
Scott MacKenzie
il 9 Lug 2021
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)
Più risposte (1)
Viranch Patel
il 9 Lug 2021
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:)
3 Commenti
Viranch Patel
il 9 Lug 2021
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.
Vedere anche
Categorie
Scopri di più su Matrix Indexing in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!