find average in for loop
Afficher commentaires plus anciens
Please help me , i want to find average y of each incease unit pr.
for example . at pr =0 . y1 = a b c
at pr = 0.5 , y 2 = a1 b1 c1
at pr = 1 , y3 = a2 b2 c2
at pr = 1.5, y4 = a3 b3 c3
at pr = 2 , y 5 = a4 b4 c4
find average of y yavg = avg(A) Avg(B) Avg(C )
note A is sum of a , a1, a2, a3, a4,
how i can code that.
a = 2;
b = 0;
n =2 ;
pr =[b:1:a];
Pr =zeros(1,length(pr));
heght = zeros(1,3);
x = zeros(1,3);
y = zeros(1,3);
r = zeros (1,n);
c = zeros(11,3)
total = zeros(1,3);
rec = zeros(3,3)
for t=1:length(pr)
rec = pr(t);
Pr(t) = rec;
for i = 1:3
Dis = 100 /7 * (i-1);
heght = 10 - Pr(t);
x(i) = Dis;
for a = 1: n
r(a) = sqrt((20 - Pr(t)) ^ 2 + x(i) ^ 2);
end
y(i) = 100* sum(r);
end
end
1 commentaire
KSSV
le 10 Juin 2020
Not clear.
Réponses (1)
John McDowell
le 10 Juin 2020
Modifié(e) : John McDowell
le 11 Juin 2020
Hi Travis,
If I'm interpretting your question correctly, I think you want to make a list of average values within each time step of your loop (assuming t is time). You're really close:
for t = 1:Arbitrary
y(t,:) = [a(t), b(t), c(t)];
end
yavg = [mean(y(:,1), mean(y(:,2), mean(y(:,3)]; % Or whatever average you choose.
I also notice that in your description, you want your time step to go up by 0.5. In order to do that, you need to ammend your pr definition to:
pr =[b:0.5:a];
I hope this helps.
Cheers,
John
2 commentaires
Travis nguyen
le 10 Juin 2020
John McDowell
le 11 Juin 2020
Hi Travis,
I'm afraid that depends on your application! I interpretted your y1, y2, y3 notication as being at time steps 1,2,3 of your loop:
y(1,:) = [a(1), b(1), c(1)];
% Then as your time step increases to 2...
y(2,:) = [a(2), b(2), c(2)];
If this is not the case, can I ask you to try rephrasing your question? There's a lot of ambiguity in there. I'm sure it's obvious to you, but for us seeing it on this end, there's a lot of ways your question could be interpretted.
Catégories
En savoir plus sur Multirate Signal Processing dans Centre d'aide et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!