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I'm comparing Loess with the Loess.jl package and results differ a bit. Steps to reproduce:
import DataInterpolations, Loess t = sort(10 .* rand(100)); u = sin.(t) .+ 0.5 * randn(100); l1 = DataInterpolations.Loess(u, t, 2, 0.75); l2 = Loess.loess(t, u, degree=2, span=0.75); using Plots scatter(t, u, label="") plot!(t, l1.(t), label="DataInterpolations.Loess") plot!(t, Loess.predict(l2, t), label="Loess.loess")
Concerning performance, it's a bit puzzling as fitting here is much faster than Loess.jl and prediction is much slower:
julia> using BenchmarkTools julia> @btime DataInterpolations.Loess($u, $t, 2, 0.75); 1.706 μs (4 allocations: 4.31 KiB) julia> @btime Loess.loess($t, $u, degree=2, span=0.75); 4.058 ms (77694 allocations: 3.82 MiB)
But performance on the prediction degrades when the data is a bit bigger:
julia> t = sort(10 .* rand(1000)); julia> u = sin.(t) .+ 0.5 * randn(1000); julia> l1 = DataInterpolations.Loess(u, t, 2, 0.75); julia> l2 = Loess.loess(t, u, degree=2, span=0.75); julia> @btime $l1.($t); 43.769 ms (17003 allocations: 94.37 MiB) julia> @btime Loess.predict($l2, $t); 4.545 ms (48492 allocations: 1.48 MiB)
The text was updated successfully, but these errors were encountered:
@piever Thanks for reporting this, Right now, I'm having my final exams, will look into it after some days.
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I'm comparing Loess with the Loess.jl package and results differ a bit. Steps to reproduce:
Concerning performance, it's a bit puzzling as fitting here is much faster than Loess.jl and prediction is much slower:
But performance on the prediction degrades when the data is a bit bigger:
The text was updated successfully, but these errors were encountered: