Stefan, You wanted to know how the nuclear option worked in comparison to usage of sum(sub(A,...) for my problem. This is just AMAZING!
@time for your 2nd suggestion i.e., sum,sub gave a time of 12.9 seconds @time for your 3rd suggestion i.e., -nuclear suggestion gave a time of 0.36. This is unbelievable!! Am I doing this right? I just took both your code, inserted into my code and only timed this specific section. Does this mean that sum and sub together take nearly 35 times long to run through 1440*1782 loops or a delay of ~40 microsecond per loop? WOW! On Wednesday, January 29, 2014 12:59:27 PM UTC-5, Stefan Karpinski wrote: > > This sum(sum(foo,2)) business is really wasteful. Just do sum(foo) to take > the sum of foo. It's also better to extract the dimensions into individual > variables. Something like this: > > function runave1(S,A,f) > s1, s2 = size(A) > p1 = f+1 > for n = f+1:s2-f-1, m = f+1:s1-f > S[m,n+1] = S[m,n] + sum(A[m-f:m+f,n+p1]) - sum(A[m-f:m+f,n-f]) > end > S > end > > > I suspect that since Matlab forces this sum(sum(X,2)) idiom on you, it > probably detects it and automatically does the efficient thing. It's > unclear to me why you need two sum operations when the slices you're taking > are just single columns, but maybe I'm missing something here. > > Currently, taking array slices in Julia makes a copy, which is > unfortunate, but in the future they will be views. In the meantime, you > might get better performance by explicitly using views: > > function runave2(S,A,f) > s1, s2 = size(A) > p1 = f+1 > for n = f+1:s2-f-1, m = f+1:s1-f > S[m,n+1] = S[m,n] + sum(sub(A,m-f:m+f,n+p1)) - sum(sub(A,m-f:m+f,n-f)) > end > S > end > > > And, of course, there's always the nuclear option for really performance > critical code, which is to write out the summation manually: > > function runave3(S,A,f) > s1, s2 = size(A) > p1 = f+1 > for n = f+1:s2-f-1, m = f+1:s1-f > t = S[m,n] > for k = m-f:m+f; t += A[k,n+p1] - A[k,n-f]; end > S[m,n+1] = t > end > S > end > > > Not so elegant, but probably the fastest possible version. Ideally, once > array slices are views, the simpler version of the code will be essentially > equivalent to this. It will take some compiler cleverness, but it's > certainly doable. It would be interesting to hear how each of these > versions performs on your data. > > On Wed, Jan 29, 2014 at 12:30 PM, John Myles White > <johnmyl...@gmail.com<javascript:> > > wrote: > >> Can you show the call to @time / @elapsed so we know exactly what's being >> timed? >> >> -- John >> >> On Jan 29, 2014, at 9:28 AM, Rajn <rjngr...@gmail.com <javascript:>> >> wrote: >> >> > Now it takes even longer i.e., ~1 minute >> > >> > Does this make sense. Also I am running this loop only once. I do not >> understand why writing in the function form would help. I read the manual >> but they suggest writing function form for something which is used many >> times. >> > I=runave(S,A,f) >> > showim(I); >> > >> > function runave(S,A,f) >> > imsz=size(A); >> > p1=f+1; >> > for n=(f+1):(imsz[2]-f-1) >> > for m=(f+1):(imsz[1]-f) >> > >> S[m,n+1]=S[m,n]+sum(sum(A[m-f:m+f,n+p1],2))-sum(sum(A[m-f:m+f,n-f],2)); >> > end >> > end >> > S; >> > end >> > >> > Do I have to declare function parameters to speed it up. >> >> >
