Template Expressions issue

[fcogneaux]

Hello!

The expressions I am looking for are of the form:
greater(X, Y)

(X and Y can be anything, a constant, a unary function, a binary function, a combination of the preceding... the all being expected to return 0 or 1)

I defined the following template expression:
template = TemplateExpressionSpec(
function_symbols=["f", "g"],
combine="((; f, g), (x1, x2, x3, x4)) -> greater(f(x1, x2), g(x3, x4))"
)

But I get the following error:
JuliaError: MethodError: no method matching isless(::ValidVector{Vector{Float64}}, ::ValidVector{Vector{Float64}})
The function isless exists, but no method is defined for this combination of argument types.

(I have tried with (x, y) -> x > y ? 1 : 0 instead of greater(x, y), but got the same error...)

Any advice? Thanks!

[MilesCranmer]

Hey, sorry, I didn't yet add the > operator for TemplateExpressions: https://github.com/MilesCranmer/SymbolicRegression.jl/blob/9fabc303dd33c624739e759d960374c7f85e56f7/src/ComposableExpression.jl#L246-L249.

For now you can manually unpack it to vector form and then wrap it up as a ValidVector again:

((; f, g), (x1, x2, x3, x4)) -> begin
    f_x = f(x1, x2)
    g_x = g(x3, x4)

    raw = [fxi > gxi ? one(fxi) : zero(fxi) for (fxi, gxi) in zip(f_x.x, g_x.x)]
    ValidVector(raw, f_x.valid && g_x.valid)
end

[MilesCranmer]

Once I get a chance, I'd like to add an loss_function_expression that allows you to specify a loss function in a way that is compatible with TemplateExpression. That would probably be the cleanest way to solve this. But it might take at least a few weeks for this to be ready. I'll try to comment here when it's done.

[MilesCranmer]

Hey @fcogneaux I got it done early; it's here: #823

[fcogneaux]

Thanks a lot Miles, I'll test it ASAP!

[MilesCranmer]

The fix is live on PySR v1.4.0