create a vector whose elements depend on its previous elements without a loop

조회 수: 12 (최근 30일)
As loops take considerable time, I'm trying without loops.
I want to create a vector whose elements depend on its previous elements, for example:
a(i) = 3*a(i-1) + 2
a(1) = 5
where a(i) is the i-th value of vector A.
Can it be done?
  댓글 수: 6
Guillaume 2019년 7월 8일
filter can't be used for that and I doubt there's anything faster than a loop.
Jan 2019년 7월 8일
This might be 10% faster than fillwithfilter:
function a = fillwithfilter2(nelem)
q = zeros(1, nelem);
q(1) = 5;
q(2:nelem) = 10;
a = filter(1, [1, -3], q);
But this is of academic interest only, if the loop is much faster.

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Guillaume 2019년 7월 5일
편집: Guillaume 2019년 7월 5일
Doing this will a loop shouldn't be slow as long as you preallocate the array beforehand:
a = zeros(1, 100); %preallocation
a(1) = 5;
for i = 2:numel(a)
a(i) = 3 * a(i-1) + 2;
It can indeed be done without a loop, with the filter function, the most difficult is to figure out the inputs. The more about section is the most helpful.
a = filter(1, [1 -3], [5, 2*ones(1, 99)]); % 1*y(n) = 1*x(n) - (-3)*y(n-1), with x = [5, 2, 2, ...] and y(0) = 0
Frankly, the loop is clearer so you should prefer it.
  댓글 수: 4
Guillaume 2019년 7월 5일
Turns out that filter is actually about 3 times slower than a loop (R2019a)
Jan 2019년 7월 8일
편집: Jan 2019년 7월 8일
@Guillaume: Even if we do something wrong (a kind of pre-mature optimization), CPUs are deterministic machines and an arithmetic operation, which needs less cycles, should consume less time.
x = ones(1, 1e6);
timeit(@() x * 2) % 1.2651e-04
timeit(@() x + 1) % 1.2923e-04
By the way, the timings are equivalent for x * 2.22 and x + 2.22, so this is not a magic effect of the JIT performing some smart repelacement for the multiplication by 2.
Nevertheless, this does not concern the question of this thread. I'm going to discuss this somewhere else.

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