Why is arrayfun for GPU slower than normal operations
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Theron FARRELL
le 28 Mai 2019
Commenté : Theron FARRELL
le 3 Juin 2019
Hi there,
Here goes a piece of testing code, yet arrayfun runs more slowly. Any thoughts? Many thanks.
function Test_GPU1()
EP = gpuArray(eps*ones(10000, 1, 'single'));
ONE = gpuArray(ones(10000, 1, 'single'));
ZERO = gpuArray(zeros(10000, 1, 'single'));
Cur_FF_Output = gpuArray(0.5*ones(10000, 1, 'single'));
Cur_Desired_Output = gpuArray(0.5*ones(10000, 1, 'single'));
for iter = 1:1000
% In output layer, Cur_Delta = Del(C)/Del(z) = Del(C)/Del(a) * Del(a)/Del(z)
% [~, Cur_Delta0] = Cost_Function_GPU(Cur_FF_Output, Cur_Desired_Output, Hyper_Para);
temp00 = Cur_FF_Output + eps;
temp11 = log(temp00);
temp22 = log(1-Cur_FF_Output+eps);
temp33 = Cur_Desired_Output.*temp11;
temp44 = 1-Cur_FF_Output.*temp22;
Cur_Delta = Cur_FF_Output-Cur_Desired_Output;
Cost = 0-sum(temp33+temp44);
temp00 = arrayfun(@plus, Cur_FF_Output, EP);
temp11 = arrayfun(@log, temp00);
temp22 = arrayfun(@log, arrayfun(@minus, ONE, arrayfun(@plus, Cur_FF_Output, EP)));
temp33 = arrayfun(@times, Cur_Desired_Output, temp11);
temp44 = arrayfun(@minus, ONE, arrayfun(@times, Cur_FF_Output, temp22));
Cur_Delta = arrayfun(@minus, Cur_FF_Output, Cur_Desired_Output);
Cost = arrayfun(@minus, ZERO, sum(temp33+temp44));
end
end
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/221446/image.jpeg)
1 commentaire
Jan
le 28 Mai 2019
Of course arrayfun has a certain overhead. It is expected to run slower than calling the operators directly.
Réponse acceptée
Joss Knight
le 28 Mai 2019
You are misunderstanding the use of arrayfun for gpuArray. Combine all those operations into a single function.
temp00 = arrayfun(@plus, Cur_FF_Output, EP);
temp11 = arrayfun(@log, temp00);
temp22 = arrayfun(@log, arrayfun(@minus, ONE, arrayfun(@plus, Cur_FF_Output, EP)));
temp33 = arrayfun(@times, Cur_Desired_Output, temp11);
temp44 = arrayfun(@minus, ONE, arrayfun(@times, Cur_FF_Output, temp22));
Cur_Delta = arrayfun(@minus, Cur_FF_Output, Cur_Desired_Output);
Cost = arrayfun(@minus, ZERO, sum(temp33+temp44));
becomes
function Cur_Delta = stuff(Cur_FF_Output, Cur_Desired_Output, EP)
temp00 = Cur_FF_Output + EP;
temp11 = log(temp00);
temp22 = log(1 - (Cur_FF_Output + EP));
temp33 = Cur_Desired_Output .* temp11;
temp44 = 1 - (Cur_FF_Output .* temp22);
Cur_Delta = Cur_FF_Output - Cur_Desired_Output;
end
Cur_Delta = arrayfun(@stuff, Cur_FF_Output, Cur_Desired_Output, EP);
Obviously, this can be extremely simplified. I've made a start, by removing the unnecessary ONE and ZERO variables.
After this, question whether you really need arrayfun, or should just call this function directly? MATLAB uses some clever optimisations that, for most sequences of element-wise operations, make using arrayfun unnecessary.
7 commentaires
Joss Knight
le 1 Juin 2019
I don't know how recently you viewed the latest documentation on GPU support. I think it's pretty comprehensive, and doesn't encourage you to use arrayfun unnecessarily. There's a great page that talks you through the various options for optimising your code which mentions arrayfun only as an advanced manoeuvre that might, but won't always, improve your performance. Generally we think of arrayfun as the way to 'write custom kernels in the MATLAB language', which many advanced users may look for. We rarely find it useful to document exactly what optimizations MATLAB is using. Usually it just confuses people, who start looking for performance improvements in the wrong place, or start blaming unexpected behaviour on the optimisations. Best to just do our best and leave ourselves the wiggle room to change the way things work from one release to the next.
Plus de réponses (1)
Jan
le 28 Mai 2019
Modifié(e) : Jan
le 28 Mai 2019
Of course arrayfun has a certain overhead. It is expected to run slower than calling the operators directly with arrays as inputs. In addition, in
Cur_FF_Output + eps
the second operand is a scalar, while in
arrayfun(@plus, Cur_FF_Output, EP)
Matlab has to process a vector. Addressing the elements of an array needs to access memory using a loop. Accessing a scalar is much cheaper.
What is the purpose of:
arrayfun(@minus, ZERO, sum(temp33+temp44))
? This is faster:
-sum(temp33+temp44)
4 commentaires
Jan
le 31 Mai 2019
Even arrayfun(@minus, 0, sum(temp33+temp44)) is too complicated compared to
-sum(temp33+temp44)
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