Ok issue settled! Thanks guys, that really helped. Actually what Jacob suggested may have already worked i.e., evaluating the code inside the function.
The reason I was still (i.e., despite function usage) getting long processing time was because instead of erasing my earlier code I decide to block comment it. I found on Google that some one suggested a way to block comment was to use ### ###. I was misled by it. The old code still hung around and was getting evaluated. I had to comment each line. I haven't still figured out a way to block comment on Julia. Maybe it does not exist? Anyway, now I actually implemented sum(sub(A...)) as Stefan suggested and implemented a function call and timed it using @time. It is coming out 13.8 seconds. I should time Matlab, but I am quite excited already! I know it comes very close. Finally a match to Matlab. Hopefully I will remain a happy camper at Julia (still struggling with Xubuntu installation though). How delighted I am. Thanks folks(developers) for your great work. A general question: I find the main Julia document very sparse in description. For example I just couldn't figure out how to use @time. The standard library just gives one line definition. I generally end up searching in the forum and most of the time I am lucky. The forum is gracious but I do not want to often post stupid questions. On Wed, Jan 29, 2014 at 6:27 PM, Rajan Gurjar <rjngrj2...@gmail.com> wrote: > Stefan, > sum(foo) did not work for me. It gave error and then I realized why. > sum(Array) turns out to be an array of (Float64,1) while the left hand > side is an element of an array. Therefore, I had to do another sum on that > one element array to give me a scalar. I just could not figure out the > error for almost an hour and then when I did > (sum(foo)) > it would give me a number but > S(i,j)=sum(foo) did not and gave error (I do not recall the error but I > think it was something like "conversion for this type not possible"!) > Of course being fresh to Julia I did not figure this out and then I > realized that this occurs usually in scipy too. > But you are correct, I will change the sum(sum) provided I am able to > figure out how to convert the array to a number. > > Also I tried plain summation without using sum just like your last section > but it is was twice slower. I have that code too and > I will recheck again to see how it differs from your last section. > > John, I will implement some of Stefan's suggestion and run the timing that > you suggested. But what I am doing is very simple. I copy paste each > section of the code (both Matlab and Julia) and run them and time them. I > find that only that last piece which I wrote takes full 40 odd seconds. But > I should find out which line takes up that time. > > Thanks > > > > > On Wed, Jan 29, 2014 at 12:59 PM, Stefan Karpinski > <ste...@karpinski.org>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 < >> johnmyleswh...@gmail.com> 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 <rjngrj2...@gmail.com> 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. >>> >>> >> >
