sc = 
Is it possible to make this tiny loop faster?
显示 更早的评论
It seems that it might be possible to make this loop faster. Does anyone have any thoughts? I have to call a loop like this millions of times in a larger workflow, and it is getting to be the slowest part of the code. I appreciate any insights!
a = 2; % constant
b = 3; % constant
n = 10; % number of elements of outputs c and s
% n could be up to 100
% preallocate and set the initial conditions
c = zeros(n,1);
s = c;
c(1) = 3; % set the initial condition for c
s(1) = 1; % set the initial condition for s
% run the loop
for i = 2:n
c(i) = a*c(i-1) -b*s(i-1);
s(i) = b*c(i-1) + a*s(i-1);
end
2 个评论
Image Analyst
2025-5-16
I ran that loop a million times with n=100 and it took only 0.33 seconds. How long does it take for you? How fast do you need it to be.
Also, what physical process is this supposed to simulate? I'm wondering because some of the values get up to 10^55 which is extremely large for any real world situation.
Christopher Smith
2025-5-17
采纳的回答
This saves considerable time for large n, but for small n, all bets are off.
a = .2; % constant
b = .3; % constant
n = 1e6; % number of elements of outputs c and s
x=[3;1];
timeit( @() loopMethod(a,b,x,n) )
timeit( @() vectorizedMethod(a, b, x,n))
function results = loopMethod(a,b,x,n)
% x=[c1;s1]
%
%results=[c1,c2,c3,....cn;
% s1,s2,s3,....sn]
A = [a, -b; b, a];
results=nan(2,n);
results(:,1)=x;
tmp=x;
for k = 2:n
tmp=A*tmp;
results(:,k)=tmp;
end
end
function results = vectorizedMethod(a, b, x,n)
% x=[c1;s1]
%
%results=[c1,c2,c3,....cn;
% s1,s2,s3,....sn]
z = a + 1i * b; % complex form of A
w = x(1) + 1i * x(2); % complex form of vector x
zAng=angle(z);
zMag=abs(z);
wAng=angle(w);
wMag=abs(w);
k = 1:n-1;
yMag=wMag*zMag.^k;
yAng=wAng+zAng.*k;
results = [x(1) yMag.*cos(yAng); x(2) yMag.*sin(yAng)]; % convert to 2xN real matrix
end
更多回答（1 个）
Akira Agata
2025-5-16
移动：Voss
2025-5-16
2 个投票
If you need to calculate this efficiently, I believe the following relationship can be utilized.
7 个评论
Did I do it correctly? It doesn't seem to give the same answers. And it's 69 times slower.
a = 2; % constant
b = 3; % constant
n = 10; % number of elements of outputs c and s
% n could be up to 100
% preallocate and set the initial conditions
c = zeros(n,1);
s = c;
% Preallocate to speed up
c = zeros(1, n);
s = zeros(1, n);
c(1) = 3; % set the initial condition for c
s(1) = 1; % set the initial condition for s
% run the loop
tic
for k = 1 : 100000
for i = 2:n
c(i) = a*c(i-1) -b*s(i-1);
s(i) = b*c(i-1) + a*s(i-1);
end
end
t1 = toc
c
s
% Method 2
% Preallocate to speed up
ct = zeros(1, n);
st = zeros(1, n);
ct(1) = 3; % set the initial condition for c
st(1) = 1; % set the initial condition for s
tic
for k = 1 : 100000
for i = 2:n
T = [a, -b; b, a] .^ (i - 1);
m = T * [ct(1); st(1)];
ct(i) = m(1);
st(i) = m(2);
end
end
t2 = toc
ratio = t2/t1
ct
st
Looks like it works to me.
syms a b
syms c0 s0
c(1) = c0;
s(1) = s0;
N = 15;
for i = 2 : N
c(i) = a*c(i-1) -b*s(i-1);
s(i) = b*c(i-1) + a*s(i-1);
end
sc = expand(c(end))
ss = expand(s(end))
t2 = [a -b; b a]^(N-1) * [c(1); s(1)];
st2 = expand(t2(1));
isequal(sc, st2)
ss2 = expand(t2(2));
isequal(ss, ss2)
Hmmm, looping does appear to be faster, at least for moderate n (up to at least 50000).
c_init = 1;
s_init = 0.5;
a = 0.2; % constant
b = 0.3; % constant
n = 500; % number of elements of outputs c and s
c = zeros(1,n);
s = zeros(1,n);
c(1) = c_init;
s(1) = s_init;
tic
for i = 2 : n
c(i) = a*c(i-1) -b*s(i-1);
s(i) = b*c(i-1) + a*s(i-1);
end
toc
tic
t = [a -b; b a]^(n-1) * [c(1); s(1)];
toc
c(end) - t(1)
s(end) - t(2)
c_init = 1;
s_init = 0.5;
a = 0.2; % constant
b = 0.3; % constant
n = 500; % number of elements of outputs c and s
M = [a,-b;b,a]^(n-1); % constant
c = zeros(1,n);
s = zeros(1,n);
c(1) = c_init;
s(1) = s_init;
tic
for i = 2 : n
c(i) = a*c(i-1) -b*s(i-1);
s(i) = b*c(i-1) + a*s(i-1);
end
toc
tic
t = M * [c(1); s(1)];
toc
Walter Roberson
2025-5-17
I think M = [a,-b;b,a]^(n-1) is taking a notable amount of time.
Maybe the analytical solution of the recurrence can save time - I didn't check it.
I assumed that the complete vectors c(1:n) and s(1:n) are required. If only c(n) and s(n) are needed, things will become much faster, of course.
c_init = 3;
s_init = 1;
a = 2; % constant
b = 3; % constant
n = 10; % number of elements of outputs c and s
F = cumprod([1;(a + 1i*b)*ones(n-1,1)]);
% N = (0:(n-1)).';
% F = (a + 1i*b).^N;
reF = real(F);
imF = imag(F);
c = c_init*reF - s_init*imF
s = s_init*reF + c_init*imF
I tested it and: oh dear, that's really slow !
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