find average in for loop

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Travis nguyen
Travis nguyen il 10 Giu 2020
Modificato: John McDowell il 11 Giu 2020
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

Risposte (1)

John McDowell
John McDowell il 10 Giu 2020
Modificato: John McDowell il 11 Giu 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 Commenti
Travis nguyen
Travis nguyen il 10 Giu 2020
Thanks John,
i have a question . for the y1(t,:) = [a(t), b(t), c(t)]; do i have to do for y2(t,:) and y3(t,:)
John McDowell
John McDowell il 11 Giu 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.

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