find two minimum values not followed by each other.
1 visualizzazione (ultimi 30 giorni)
Mostra commenti meno recenti
I am using mink(A,2) to find the two smallest values, but I need to find the two smallest values not followed by each other. How can this be done?
To illustrate: A(1; 1.2; 1.2; 0.7; 0.6; 0.61; 1; 1; 1.2;0.65) I want to find 0.6 and 0.65 and not 0.6 and 0.61.
Risposta accettata
Ameer Hamza
il 19 Apr 2020
Modificato: Ameer Hamza
il 20 Apr 2020
Try this. It also preserve the order of minumum values
A = [0.65;1.2;1.2;0.7;0.6;0.61;1;1;1.2;1];
[vals,idx] = mink(A,3);
if diff(idx(1:2))==1
I = sort(idx([1 3]));
min_vals = A(I);
else
I = sort(idx([1 3]));
min_vals = A(I);
end
Result:
I =
1
5
min_vals =
0.6500
0.6000
2 Commenti
Più risposte (2)
Mehmed Saad
il 19 Apr 2020
Modificato: Mehmed Saad
il 19 Apr 2020
The long Method
A=[1;1.2;1.2;0.7;0.6;0.61;1;1;1.2;0.65];
miny = [];
for i =1:3
x = min(A);
A(A==x) = [];
if(i~=2)
miny = [miny x];
end
end
miny
The short one
A=[1;1.2;1.2;0.7;0.6;0.61;1;1;1.2;0.65];
B = sort(A);
miny = [B(1) B(3)]
3 Commenti
Image Analyst
il 19 Apr 2020
So the long method is too long, but the short method is too short. Exactly how many lines of code do you want?
Image Analyst
il 19 Apr 2020
Try this:
A=[1;1.2;1.2;0.7;0.6;0.61;1;1;1.2;0.65];
% Plot, just for fun.
plot(A, 'r.-', 'MarkerSize', 30);
grid on;
% Find the values in ascending order, so the min value will be at location/index = 1.
[sortedA, sortOrder] = sort(A, 'Ascend')
% Now find the second lowest value but it can't be in the location next to the lowest value.
% Use diff() to see how many indexes the values are separated by each other by.
d = diff(sortOrder) % Don't want 1's
% Get min1 and min2 according to the non-consecutive criteria.
min1 = sortedA(1)
if d(1) ~= 1
% Not consecutive so we're OK.
min2 = sortedA(2) % Get the next one
else
% Consecutive so pick the next location with the next lowest value.
min2 = sortedA(3)
end
If it's homework, don't turn in my work as your own or you could get into trouble with your instructor.
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!