Why is this line so inefficient
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I have a very simple equation that I'm coding in MATLAB that I wouldn't think should be taking as much time as it is. The initial line is in a for loop so I vectorized the problem and that turned out to be even slower (which I was pretty surprised by). Does anyone see anything obvious that I'm not considering for why this is so slow? Typically values for n and m are between 0-10
% Rnm = zeros(size(rho));
% MaximumExponent = (n-m)/2;
% s = 0:MaximumExponent;
% numerator = (-1) .^ s .* factorial(n-s);
% denominator = factorial(s) .* factorial((n+m) / 2-s) .* factorial((n-m) / 2-s);
% Rnm = sum((numerator ./ denominator) .* rho .^ (n-2 .* s), 2);
Rnm2 = zeros(size(rho));
for s=0:(n-m)/2
numerator = (-1)^s * factorial(n-s);
denominator = factorial(s)*factorial((n+m)/2-s)*factorial((n-m)/2-s);
Rnm2 = Rnm2 + (numerator / denominator) * rho.^(n-2*s);
end % for s statement
Here is a profile for the runtime...
EDIT: I should probably also add a couple more details on some things. rho will be a column vector of ~500,000 elements. I imagine the size of rho is alone the reason that most of this takes so long, but it still seems a bit abnormally long to me.
EDIT 2: Should probably include my system details...running Windows 10 Pro, 11th Gen Intel Core i5-1135G7 @ 2.4GHz, 16GB RAM. Nothing to fantastic, but not the worst PC in the world. That's why I'd think this should be running faster...
Risposte (1)
David Hill
il 26 Lug 2022
tic;
rho=rand(500000,1);n=9;m=3;
s=0:(n-m)/2;
numerator = (-1).^s .* factorial(n-s);
denominator = factorial(s).*factorial((n+m)/2-s).*factorial((n-m)/2-s);
Rnm=sum((numerator./denominator).*rho.^(n-2*s),2);
toc;
4 Commenti
David Hill
il 27 Lug 2022
tic;
rho=rand(500000,1);n=9;m=3;
f=factorial(0:n);%could do lookup table for factorials (might help speed slightly)
s=0:(n-m)/2;
numerator = (-1).^s .* f(n-s+1);
denominator = f(s+1).*f((n+m)/2-s+1).*f((n-m)/2-s+1);
Rnm=sum((numerator./denominator).*rho.^(n-2*s),2);
toc;
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