a question on for loop statement

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ektor 2019-5-16

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评论： Luna 2019-5-17

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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?

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Adam Danz 2019-5-16

编辑：Adam Danz 2019-5-16

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 2019-5-16

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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回答（1 个）

Jos (10584) 2019-5-16

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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