Why is this so slow?

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Moe Szyslak
Moe Szyslak 2020-2-21
评论： Moe Szyslak 2020-2-21
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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采纳的回答

madhan ravi
madhan ravi 2020-2-21
Instead of loop why not simply use cumsum() ?
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Rik
Rik 2020-2-21
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.
Moe Szyslak
Moe Szyslak 2020-2-21
That's much cleaner. I see similar speedups on my machine. Thanks again.

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更多回答（1 个）

Rik
Rik 2020-2-21
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)))
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Rik
Rik 2020-2-21
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.
I didn't remember the conclusion from your question, so I looked it up.
%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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