Is it better to use sqrt or ^(1/2) ?
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Andre Ged
el 15 de Mzo. de 2016
Comentada: Walter Roberson
el 8 de Sept. de 2023
In order to have better performance in terms of computational time, is it better to use sqrt or ^(1/2) ?
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KSSV
el 15 de Mzo. de 2016
Editada: per isakson
el 3 de Jun. de 2017
clc; clear all
N = 10:10:100000;
t_sqrt = zeros(length(N),1) ;
t_pow = t_sqrt ;
for i = 1:length(N)
k = rand(N(i),1) ;
t1 = tic ;
k1 = sqrt(k) ;
t_sqrt(i) = toc(t1) ;
%
t2 = tic ;
k2 = k.^0.5 ;
t_pow(i) = toc(t2) ;
end
figure
plot(t_sqrt,'r') ;
hold on
plot(t_pow,'b') ;
legend('sqrt','power')
You may check yourself.....power (^) is taking less time.
2 comentarios
Walter Roberson
el 3 de Jun. de 2017
My tests suggest that sqrt() is faster, except maybe on some very small vectors.
Note: it will take a while to execute the below as it tests over a range of sizes.
N = round(logspace(2,7,500));
t_sqrt = zeros(length(N),1) ; t_sqrt1 = t_sqrt; t_pow = t_sqrt; t_pow1 = t_sqrt;
data = rand(1,max(N));
for i = 1 : length(N); k = data(1,1:N(i)); t_sqrt(i) = timeit(@() sqrt(k),0); t_pow(i) = timeit(@() k.^(1/2), 0); end
for i = 1 : length(N); k = data(1,1:N(i)); t_sqrt1(i) = timeit(@() sqrt(k),0); end; for i = 1 : length(N); k = data(1,1:N(i)); t_pow1(i) = timeit(@() k.^(1/2),0); end
plot(N,t_sqrt,'r-',N,t_sqrt1,'r--', N,t_pow,'b-', N,t_pow1, 'b--')
legend({'sqrt, combined loop', 'sqrt, separate loops', 'pow, combined loops', 'pow, separate loop'} )
I originally had only the combined loop, and smaller upper bounds, but I noticed some oddities in the timings for which the timings were sometimes shorter in synchronized ways. That suggested that some iterations might happen to execute faster than others by chance, so I decided to also run the loops independently to see whether the timing oddities stayed with the array sizes or were instead independent between the separated attempts. The tests did end up suggesting that the timing oddities were related to array sizes. Likely at the point where MATLAB starts handing over the work to BLAS or MLK or LINPACK or whatever, the iterations get relatively faster.
Más respuestas (1)
Amarpreet
el 8 de Sept. de 2023
Both these operations result in different values when operated on a matrix, so that may also be considered while using the functions.sqrt gives square root of each element, while the pow function operates on the whole matrix.
2 comentarios
Walter Roberson
el 8 de Sept. de 2023
N = 5:5:100;
t_sqrt = zeros(length(N),1) ; t_sqrt1 = t_sqrt; t_pow = t_sqrt; t_pow1 = t_sqrt;
data = rand(max(N));
for i = 1 : length(N); k = data(1:N(i),1:N(i)); t_sqrt(i) = timeit(@() sqrtm(k),0); t_pow(i) = timeit(@() k^(1/2), 0); end
for i = 1 : length(N); k = data(1:N(i),1:N(i)); t_sqrt1(i) = timeit(@() sqrtm(k),0); end; for i = 1 : length(N); k = data(1:N(i),1:N(i)); t_pow1(i) = timeit(@() k^(1/2),0); end
plot(N,t_sqrt,'r-',N,t_sqrt1,'r--', N,t_pow,'b-', N,t_pow1, 'b--')
legend({'sqrt, combined loop', 'sqrt, separate loops', 'pow, combined loops', 'pow, separate loop'} )
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