I would urge you to reconsider the usage of cell arrays. When you index into a cell array, you're telling the compiler that it cannot safely make any assumptions about what's going to come out of the array, which may cause the compiler to box values that could otherwise be unboxed. Assuming that such unnecessary boxing occurs, it could potentially cause superfluous memory allocation.
-- John On Dec 1, 2014, at 7:08 PM, WuYijing <[email protected]> wrote: > Hi John, > > Actually I want to put 0-1000 into a n by n matrix, and I want each entry to > be a array containing some of these numbers from 1 to 1000. Given an pair > (i,j), I want to get an array B[i,j], so I think cell is a better way to do > it than putting arrays in arrays (like [[],[],…]). > > And I want to use B[2,3][i] to calculate something, and it seems better if I > use the second method(creating a list T rather than use x), when i bigger > than 512, in first method, both 2 steps requires allocations, but in the > second one, the second @time doesn’t require allocation. > > @time x=B[2,3][i] > @time y=x^2+2*x+sqrt(x) > println(x) > > > @time T[i]=B[2,3][i] > @time y=T[i]^2+2*T[i]+sqrt(T[i]) > println(T[i]) > > This problem really frustrated me, and I really appreciate your advice. The > code is for your convenience: > function test() > > B=fill!(cell(5,5),Int64[]) > > for i=1:513 > > push!(B[2,3],i) > > end > > T=zeros(Int64,513) > > y=0.0 > > for i=1:513 > > @time T[i]=B[2,3][i] > > @time y=T[i]^2+2*T[i]+sqrt(T[i]) > > println(T[i]) > > end > > for i=1:513 > > @time x=B[2,3][i] > > @time y=x^2+2*x+sqrt(x) > > println(x) > > end > > > > end > > > Yijing > > >> On Dec 1, 2014, at 8:49 PM, John Myles White <[email protected]> >> wrote: >> >> Hi Yijing, >> >> Thanks for the update. I'm a little confused by somethings happening in your >> code. >> >> (1) Do you really need to use cell(5, 5)? In general, you'll want to use >> more tightly typed arrays. >> >> (2) Do you really want to fill the B array with many copies of the exact >> same pointer? This seems improbable. Did you perhaps want to create a new >> Int64[] array for every element of B? >> >> -- John >> >> On Dec 1, 2014, at 6:45 PM, Yijing Wu <[email protected]> wrote: >> >>> Sorry it seems that I simplified my problem too much in the previous code. >>> Actually I am running something similar to the following, and I run it as a >>> function and it starts to have allocation since i=512. I am really unsure >>> if it's my coding problem(so it can be improved to 0 bytes) or it is >>> natural in the system that allocations appear. Thank you so much! >>> >>> function test() >>> B=fill!(cell(5,5),Int64[]) >>> >>> for i=1:513 >>> push!(B[2,3],i) >>> end >>> >>> T=zeros(Int64,513) >>> >>> for i=1:513 >>> @time T[i]=B[2,3][i] >>> println(T[i]) >>> end >>> end >>> >>> >>> 在 2014年12月1日星期一UTC-6下午7时24分31秒，John Myles White写道： >>> Did you run this inside a function? If not, your results are not going to >>> be useful indicators of how code will perform inside a function. >>> >>> Inside of a function body, I see 0 bytes being allocated. >>> >>> -- John >>> >>> On Dec 1, 2014, at 4:53 PM, Yijing Wu <[email protected]> wrote: >>> >>>> Hi all, I found a strange problem about a very simple code in julia and >>>> hopefully I can get some help from you, here is the code: >>>> >>>> >>>> B=[1:1000] >>>> T=zeros(Int64,1000) >>>> for i=1:1000 >>>> @time T[i]=B[i] >>>> println(T[i]) >>>> end >>>> >>>> And when I run the code, the @time shows that it require 48 bytes >>>> allocation when i is larger than or equal to 512, and 0 bytes when >>>> smaller. Is this a problem that can be improved or I have to accept that >>>> it is designed to take some allocations when larger than 512? >>>> >>>> Thanks a lot for your help! >>>> >>> >> >
