Custom Expression with Dimensional Consistency

[RaymondWKWong]

I'm running into an issue when trying to use a custom expression (via expression_spec) together with dimensionally consistent units (via X_units and y_units) in PySR. In my code, everything works as expected when I either:

• Do not define X_units and y_units but define expression_spec
• Define X_units and y_units but without defining expression_spec

However, when I try to combine both (i.e., use a custom expression and set X_units and y_units), the code fails. My main questions are:

1. What might be causing the conflict between a custom expression and the enforcement of dimensional consistency?
2. If I know the units for the constants (for example, if f is dimensionless, but constants like y0_A, k_A, y0_M, k_M have known physical units), is there a way to incorporate these unit definitions during the search process?

Below is a simplified snippet of my code for context:

template_expression = TemplateExpressionSpec(
    function_symbols=["f", "y0_A", "k_A", "d_A", "y0_M", "k_M", "d_M"],
    combine=("((; f, y0_A, k_A, d_A, y0_M, k_M, d_M), "
             "(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11)) -> "
             "abs(f()) * (abs(y0_M()) + abs(k_M()) * (d_M(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11))^(-0.5)) + "
             "(1 - abs(f())) * (abs(y0_A()) + abs(k_A()) * (d_A(x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11))^(-0.5))")
)

model = PySRRegressor(
    binary_operators=["+", "-", "*", "/"],
    maxsize=100,
    maxdepth=30,
    niterations=500,
    population_size=100,
    expression_spec=template_expression,  # Using the custom expression
    dimensionless_constants_only=False,
    dimensional_constraint_penalty=10000
)

model.fit(
    x_train,
    y_train,
    X_units=["J/s", "mm/s", "μm", "μm", "μm", "J/m^3", "1", "1", "1", "K/s", "K*s/m"],
    y_units="Pa"
)

Thank you!

[MilesCranmer]

Unfortunately, template expression specifications are currently incompatible with dimensional constraints, apologies. There should certainly be a better error message though.

[RaymondWKWong]

That makes sense, thanks for the clarification!

Unfortunately, computing dimensionless quantities isn’t really feasible in my case, as interpretability of each input is important.

I was wondering what specifically makes template expressions incompatible with dimensional constraints? Is it a fundamental limitation in how the expressions are parsed, or is it something that could be addressed in a future update? Is there any possible workaround?

Thanks!

[MilesCranmer]

It's possible, but it will take a bit of work. I don't have plans to add it myself. If you are interested, you would need to fill in code here: https://github.com/MilesCranmer/SymbolicRegression.jl/blob/3a85b69659e387383f3210960460e4557eb259a8/src/TemplateExpression.jl#L682-L689. See https://github.com/MilesCranmer/SymbolicRegression.jl/blob/master/src/DimensionalAnalysis.jl for details.

[MilesCranmer]

Although I don't think it's possible for ALL cases, since template expressions literally allow anything in the template. It would need to constrain to only specific types of template expressions and throw errors otherwise.

[MilesCranmer]

If there’s a way to compute dimensionless quantities for your problem, that might be better to use as input for a template expression?

[alexlib]

Sorry if I detour the discussion, I know that scimed was developed to be dimensionally consistent
https://github.com/LironSimon/SciMED

[MilesCranmer]

PySR can do this, see https://ai.damtp.cam.ac.uk/pysr/examples/#10-dimensional-constraints. This particular discussion thread is about compatibility with the new TemplateExpressionSpec feature: https://ai.damtp.cam.ac.uk/pysr/examples/#template-expressions