How to select the most complex model? idx = "max"

[snowysnowcones]

Hello,
Using the Julia Symbolic Regression package. It's wonderful. Thank you everyone for your work.
Probably a dumb question, but in some instances I don't care about how complex the equation is and I just want the equation with the lowest loss. How can I predict against the most complex model?

I see in the documentation I can override the Pareto frontier equation with:
predict(mach, (data=X, idx=2))

but I'm not sure how to get the "max" idx programmatically.

Thanks for your help!

edit
After some playing around, I think I could do something like this:

r=report(mach) predict(mach,(data=X,idx=size(r.scores,1))
which should give me the maximum value of the final list of equations.

Hopefully this is the optimal way.

[MilesCranmer]

Your way is valid already.

An alternative: you can also pass selection_method which is a property of SRRegressor. So for your case you could pass:

SRRegressor(
    #= ... =#
    selection_method=(; losses, kws...) -> argmin(losses),
)

I think this should work for you. But let me know if it doesn't.

The default selection method is here: https://github.com/MilesCranmer/SymbolicRegression.jl/blob/eafea58565510cca5ed1b8f832ccf27fd233414e/src/MLJInterface.jl#L611

[snowysnowcones]

Yes, this also works, thank you!