a question on for loop statement

3 vues (au cours des 30 derniers jours)
ektor
ektor le 16 Mai 2019
Commenté : Luna le 17 Mai 2019
Dear all,
I have this for loop
T=1000;
k=0.1;
u=rand(T,1);
a = zeros(T,1);
a(1) =u(1)+ k*0.01;
for t=2:T
a(t) = u(t,1) + k*a(t-1);
end
Is there a faster way of obtaining a? Maybe if I avoid loop?
  4 commentaires
Adam Danz
Adam Danz le 16 Mai 2019
Modifié(e) : Adam Danz le 16 Mai 2019
This is the tricky part: *a(t-1)
Short answer to "is there a faster way": Probably not.
There's probably a way to avoid the loop by replacing it with a convoluted, unreadable, jumble of functions but I doubt it will be as fast and it will not be as intuitive. If your loop works for you, keep it. It's simple, clean, and fast.
Luna
Luna le 16 Mai 2019
I agree with Adam I have tried with both T = 1000 and T = 1000000.
The time perfomances are below:
T = 1000 -> Elapsed time is 0.051244 seconds.
T = 1000000 -> Elapsed time is 0.073614 seconds.
The for loop is already as fast as it could be and the simplest solution.

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Réponses (1)

Jos (10584)
Jos (10584) le 16 Mai 2019
This is filtering.
T=10; % smaller example
k=0.1;
u=rand(T,1);
% your loop -> a
a = zeros(T,1);
a(1) =u(1)+ k*0.01; % i do not get this addition ...
for t=2:T
a(t) = u(t,1) + k*a(t-1);
end
% filtering -> aa
uu = u ;
uu(1) = uu(1) + k*0.01 ; % implement offset?
aa = filter(1, [1 -k], uu) ;
% do they produce the same result?
isequal(a, aa) % YES
  9 commentaires
Jos (10584)
Jos (10584) le 17 Mai 2019
btw, regarding execution time, you should also include the pre-allocation of the array :-D
Luna
Luna le 17 Mai 2019
+1 Jos :)

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