You may want to check out Benchmarks.jl <https://github.com/johnmyleswhite/Benchmarks.jl>, which goes to significant lengths to do benchmarking correctly.
On Wed, Feb 3, 2016 at 4:05 PM, Milan Bouchet-Valat <[email protected]> wrote: > Le mercredi 03 février 2016 à 12:55 -0800, Christopher Alexander a > écrit : > > Try doing something like: > > > > tic() > > # my code > > toc() > Be careful with tic() and toc() in Julia. In most cases, when > benchmarking, you should wrap your code in a function to make sure it > gets specialized on the argument types, instead of running it from the > REPL. So in general it's better to do: > @time myfun(arg1, arg2, ...) > > See http://docs.julialang.org/en/latest/manual/performance-tips/#measur > e-performance-with-time-and-pay-attention-to-memory-allocation > > > Regards > > > On Wednesday, February 3, 2016 at 3:28:28 PM UTC-5, Lytu wrote: > > > Hello Julia users, > > > > > > Can someone tell me what's the equivalent of matlab elapsed cputime > > > in Julia > > > > > > For example i Matlab, we can do this: > > > t = cputime; > > > x=4; > > > iter = 1; > > > z = ones(1,4); > > > y=x*2*z; > > > e = cputime-t > > > > > > But in Julia i don't seem to find how to do this. I thought i can > > > use > > > t=time() > > > x=4; > > > iter = 1; > > > z = ones(1,4); > > > y=x*2*z; > > > e=time()-t > > > > > > But time() in Julia is not the elapsed CPU time, it's a wall clock > > > time. > > > > > > Can someone help me? > > > > > > Thank you > > > >
