Multiexpression Outputs Question

[ccclalala123]

When I define multiexpression outputs, how do I know which is P and which is Q?

objective = """
function my_loss_function(tree, dataset::Dataset{T,L}, options)::L where {T,L}

  tree.degree != 2 && return L(Inf)

    P = tree.l
    Q = tree.r

    # Evaluate numerator:
    P_prediction, flag = eval_tree_array(P, dataset.X, options)
    !flag && return L(Inf)

    # Evaluate denominator:
    Q_prediction, flag = eval_tree_array(Q, dataset.X, options)
    !flag && return L(Inf)

    R12_2D =  (P_prediction .- 1)  .* cz1_jl ./ (0.09 .* woo_tran_jl .* (Q_prediction .+ c_pe_tran_jl .- 1.0))
    
    diffs = R12_2D .- dataset.y
    loss = sum(diffs .^ 2) 

    return loss
 
end
"""

model = PySRRegressor(
    population_size=200,
    niterations=200,
    binary_operators=["+", "-", "*", "/"],
    loss_function=objective,
)
model.fit(X, y)
model.get_best().equation

The results are as follows：

Complexity,Loss,Equation
1,Inf,"x0"
3,1.8108e-6,"0.6919551 + 1.4003153"
5,1.6049491e-6,"(0.43112686 - x0) * 2.6639194"
7,1.3138252e-6,"(x0 / (-0.3637128 - x0)) * 4.5015717"
9,1.048514e-6,"((x0 / (-0.32570758 - x0)) + x0) / 4.502915"
11,1.0421031e-6,"(((x0 / (-0.32833982 - x0)) + x0) / 0.90415287) - 4.490343"

[MilesCranmer]

Basically .l is the left side and .r is the right side of the outermost operator. So if you were numerically evaluating the equation, the last operator that you would evaluate is the one you split at. So x0 / (-0.3637128 - x0)) * 4.5015717 has P equal to the x0/(-0.36 - x0) and Q equal to the 4.5. The outermost operator * is ignored.

But yes I think it would be nice for PySR to have a proper way of facilitating this, rather than needing to do it all manually.

[ccclalala123]

I got it! Thank you so much for your quick response!

[ccclalala123]

If there are 3 or more multi-expression outputs, how to realize and distinguish them? For example, Y=P×2+Q×3+W×4+T×5+K×6，where P, Q, W, T and K are function expressions of x.

[MilesCranmer]

Want to make a new thread for that? But basically just apply the pattern recursively

[ccclalala123]

How to achieve it in a simple way? Because I need to train 12 expressions of coefficients in an equation.

[Alexia369]

How to achieve it in a simple way? Because I need to train 12 expressions of coefficients in an equation.

Do you have the code to achieve it？ I have met the same question