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

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Galgool 2019-7-5

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评论： Jan 2019-7-8

采纳的回答： Guillaume

Hi,

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?

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Guillaume 2019-7-8

filter can't be used for that and I doubt there's anything faster than a loop.

Jan 2019-7-8

在 MATLAB Online 中打开

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);
end

But this is of academic interest only, if the loop is much faster.

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采纳的回答

Guillaume 2019-7-5

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编辑：Guillaume 2019-7-5

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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;
end

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.

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Guillaume 2019-7-5

Turns out that filter is actually about 3 times slower than a loop (R2019a)

Jan 2019-7-8

编辑：Jan 2019-7-8

在 MATLAB Online 中打开

@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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