Why is this so slow?
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Moe Szyslak
il 21 Feb 2020
Commentato: Moe Szyslak
il 21 Feb 2020
Hi, All -- the seemingly simple script shown below is exceedingly slow. It takes several minutes on a fast workstation. I can't figure out why it takes so long. What am I missing? Thanks, in advance.
N = 10^6;
a = 0;
b = 4;
r = a + (b-a)*rand(N,1);
c = r.*sinh(r);
k = (1:N)';
s(N,1) = 0;
for j = 1:N
s(j,1) = sum(c(1:j));
end
s = (b-a)*s./k;
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madhan ravi
il 21 Feb 2020
Instead of loop why not simply use cumsum() ?
3 Commenti
Rik
il 21 Feb 2020
If you want to implement it with a loop, you can still gain a lot of speed:
N=10^4;
c=rand(N,1);
tic
s=[];s(N,1)=0;
for n=1:N
s(n)=sum(c(1:n));
end
t1=toc;
tic
s=[];s(N,1)=0;
s(1)=c(1);
for n=2:N
s(n)=s(n-1)+c(n);
end
t2=toc;
tic
s=cumsum(c);
t3=toc;
clc
fprintf('fast loop is %.0f times faster\ncumsum is %.0f times faster\n',...
t1/t2,t1/t3)
That is over 100 times faster on my machine, with cumsum increasing the speed 10 times over that.
Più risposte (1)
Rik
il 21 Feb 2020
Loops with many elements are slow. You also forgot to properly pre-allocate s.
N = 10^5;
a = 0;
b = 4;
r = a + (b-a)*rand(N,1);
c = r.*sinh(r);
k = (1:N)';
%s(N,1) = 0;
s=zeros(N,1);
for j = 1:N
s(j,1) = sum(c(1:j));
end
s = (b-a)*s./k;
clc
s2 = (b-a)*cumsum(c)./k;
fprintf('largest error = %.2e\n',max(abs(s-s2)))
4 Commenti
Rik
il 21 Feb 2020
It works if the variable doesn't exist yet, but if it already exists (e.g. because you aborted a 10^6 run to do a 10^5 instead), this will cause a dimension error with this code.
%replace this
s(N,1) = 0;
%with this
s=[];s(N,1) = 0;
Because apparently that is faster.
Except in this case, where it is faster to create the array with cumsum instead.
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