Stefan, I was just playing around in Julia to compare with matlab. Here's the code, on where and why I needed linspace to return a vector. Pretty common use cases for laying a grid in space, when trying to solve PDEs
Also Julia code suggestions welcome ;) 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] <javascript:>> 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. >>> >>>
