Defining a custom operator without needing the sympy mapping

[MilesCranmer]

This was in an email discussion with a user:

How can I define a custom operator, without needing to define a sympy mapping for it?

The solution is to define a sympy.Function with the name of your custom operator - then sympy won't map it into other sympy functions.

For example, say we want to define a custom logarithm function:

import sympy
from pysr import PySRRegressor

class my_log(sympy.Function):
    pass

model = PySRRegressor(
    unary_operators=["my_log(x::T) where T = (x > zero(T)) ? log(x) : T(NaN)"], 
    extra_sympy_mappings={"my_log": my_log}
)

Then, this can be used normally, and the sympy export will just use my_log, rather than needing to convert it to native sympy functions.

This could also be used if you want to define custom LaTeX for some operator.