Effacer les filtres
Effacer les filtres

find two minimum values not followed by each other.

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FV
FV le 19 Avr 2020
Commenté : Ameer Hamza le 20 Avr 2020
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.
  1 commentaire
FV
FV le 19 Avr 2020
I also need to find the index of the values.

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Réponse acceptée

Ameer Hamza
Ameer Hamza le 19 Avr 2020
Modifié(e) : Ameer Hamza le 20 Avr 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 commentaires
FV
FV le 20 Avr 2020
Looks good, How can I get the index of the result?
Ameer Hamza
Ameer Hamza le 20 Avr 2020
See the updated code.

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Plus de réponses (2)

Mehmed Saad
Mehmed Saad le 19 Avr 2020
Modifié(e) : Mehmed Saad le 19 Avr 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 commentaires
Image Analyst
Image Analyst le 19 Avr 2020
So the long method is too long, but the short method is too short. Exactly how many lines of code do you want?
FV
FV le 19 Avr 2020
I don't want to sort it and therefore the short one doesn't work. I also need it to work for other A's.

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Image Analyst
Image Analyst le 19 Avr 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.
  1 commentaire
FV
FV le 19 Avr 2020
This code also just work for this problem. Sometimes it's no smaller values beside the smallest one.
This is not a homework.

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