How can I tell Julia what are the parameters in a linear model?

pdeffebach · September 26, 2022, 5:24pm

There are many performance improvements to be had before changing the algorithm. In particular, you want to “capture” the variables used in the optimization inside a let block.

Also, never do df.z in performant code, since DataFrames are type-unstable. Always work on the vector directly.

julia> z = df.z; x = df.x; y = df.y;

julia> g = let x = x, y = y, z = z 
           β -> begin 
                   x₀ = β[3]; y₀ = β[4]
                   ẑ =  β[1]*(x .- x₀).^2 + β[2]*(y .- y₀).^2 
                   return sum((z - ẑ).^2)
           end
       end;