Strange it *was* giving me an error saying deprecated and that I should use collect, but now it's fine.
On Tuesday, September 29, 2015 at 10:28:12 PM UTC-4, Sheehan Olver wrote: > > fez, I'm pretty sure the code works fine without the collect: when exp is > called on linspace it converts it to a vector. Though the returned t will > be linspace object. > > On Wednesday, September 30, 2015 at 12:10:55 PM UTC+10, feza wrote: >> >> Here's the code I was using where I needed to use collect (I've been >> playing around with Julia, so any suggestions on this code for perf is >> welcome ;) ) . In general linspace (or the : notation) is also used >> commonly to lay a grid in space for solving a PDE for some other use >> cases. >> >> function gp(n) >> n = convert(Int,n) >> t0 = 0 >> tf = 5 >> t = collect( linspace(t0, tf, n+1) ) >> sigma = exp( -(t - t[1]) ) >> >> c = [sigma; sigma[(end-1):-1:2]] >> lambda = fft(c) >> eta = sqrt(lambda./(2*n)) >> >> Z = randn(2*n) + im*randn(2*n) >> x = real( fft( Z.*eta ) ) >> return (x, t) >> end >> >> >> On Tuesday, September 29, 2015 at 8:59:52 PM UTC-4, Stefan Karpinski >> wrote: >>> >>> I'm curious why you need a vector rather than an object. Do you mutate >>> it after creating it? Having linspace return an object instead of a >>> vector was a bit of a unclear judgement call so getting feedback would >>> be good. >>> >>> On Tuesday, September 29, 2015, Patrick Kofod Mogensen < >>> [email protected]> wrote: >>> >>>> No: >>>> >>>> julia> logspace(0,3,5) >>>> 5-element Array{Float64,1}: >>>> 1.0 >>>> 5.62341 >>>> 31.6228 >>>> 177.828 >>>> 1000.0 >>>> >>>> On Tuesday, September 29, 2015 at 8:50:47 PM UTC-4, Luke Stagner wrote: >>>>> >>>>> Thats interesting. Does logspace also return a range? >>>>> >>>>> On Tuesday, September 29, 2015 at 5:43:28 PM UTC-7, Chris wrote: >>>>>> >>>>>> In 0.4 the linspace function returns a range object, and you need to >>>>>> use collect() to expand it. I'm also interested in nicer syntax. >>>>> >>>>>
