Thanks for the information. I just read up on the planed changes for 0.5, but I was not aware of this yesterday - so much to learn and keep yourself updated on! With regards to (A'.*B)[2:end, 2:end] I didn't really think about using this trick before, but I guess you meant (A[:,1].*B)[2:end, 2:end] right? There are two columns in A, I'm not sure I made that too clear above. I guess that would spare me the temporary from B[:, 2:end] which is the bigger of the two.
Thanks for all the good ideas, I will have to try the things out, and see what works best. On Tuesday, September 15, 2015 at 2:31:16 PM UTC-4, Stefan Karpinski wrote: > > Both the slices and the transpose create temporaries; this will change in > 0.5 but for now it's the way things work. You could do something like > (A'.*B)[2:end,2:end] to avoid some temporaries. But if you want to > eliminate all temporaries, writing out the loops is the best way at the > moment. > > On Mon, Sep 14, 2015 at 9:36 PM, Patrick Kofod Mogensen < > patrick....@gmail.com <javascript:>> wrote: > >> Thank you both. Maybe I was somewhat unclear still, but it should be >> >> >> for i in 2:n, j in 1:n >> C[j,i-1] = A[i,1] * B[j,i] >> end >> >> At least to do what I was trying to explain above. >> >> >> On Monday, September 14, 2015 at 9:21:45 PM UTC-4, Ismael VC wrote: >>> >>> Could also be with this syntax: >>> >>> >>> for i in 2:i, j in 1:j >>> C[i,j] = A[i,1] * B[j,i] >>> end >>> >>> >>> >>> El lunes, 14 de septiembre de 2015, 20:08:12 (UTC-5), Tom Breloff >>> escribió: >>>> >>>> How about you do it in a loop (note this should be done in a >>>> function... globals will ruin performance): >>>> >>>> C = Array(n,n-1) >>>> for i in 2:i >>>> for j in 1:j >>>> C[i,j] = A[i,1] * B[j,i] >>>> end >>>> end >>>> >>>> On Mon, Sep 14, 2015 at 8:45 PM, Patrick Kofod Mogensen < >>>> patrick....@gmail.com> wrote: >>>> >>>>> Sorry, something went wrong with my description. It was supposed to be: >>>>> >>>>> C =A[2:end, 1]' .* B[:, 2:end] >>>>> >>>>> A is n-by-2 and B is n-by-n (difference!). So B[:, 2:end] is >>>>> n-by-(n-1) and I expect C to be n-by-(n-1). Sorry for this mistakes! >>>>> >>>>> >>>>> >>>>> On Monday, September 14, 2015 at 8:39:59 PM UTC-4, Tom Breloff wrote: >>>>>> >>>>>> What size do you expect your result to be? Are you creating a 1 x >>>>>> (b-1) row vector? If so, I would check out the ArrayViews package and >>>>>> do >>>>>> something like: >>>>>> >>>>>> using ArrayViews >>>>>> A1 = view(A,:,1) >>>>>> C = Float64[dot(A1, view(B,:,i)) for in 2:size(B,2)] >>>>>> >>>>>> On Mon, Sep 14, 2015 at 7:27 PM, Patrick Kofod Mogensen < >>>>>> patrick....@gmail.com> wrote: >>>>>> >>>>>>> I have a line in my code that seems to really allocate a lot of >>>>>>> memory, and I think it might be slowing down my program. I have two >>>>>>> matrices. The first matrix A is n-by-2 and the second matrix B is >>>>>>> n-by-b. >>>>>>> What I need to do is to multiply the first column of A element by >>>>>>> element >>>>>>> to each row in B[:, 2:end]. I use broadcasting (.*) to achieve this >>>>>>> along >>>>>>> with a transpose of A[:,1]: >>>>>>> >>>>>>> C=A[:,1]'.*B[:,2:end] >>>>>>> >>>>>>> However, it seems to allocate a lot of temporary memory (the number >>>>>>> memory number in .mem overflows even for small problems). Is there >>>>>>> anything >>>>>>> I can do here? >>>>>>> >>>>>>> >>>>>>> P >>>>>>> >>>>>> >>>>>> >>>> >
