[BUG]: PySR can not rediscover X^2 about pysr HOT 3 CLOSED

Looks like it is not a proper place where this bug is. With the following Julia script, the same problem is:

using SymbolicRegression
using MLJ
X = 8randn(1000, 1)
y = @. X[:, 1]^2 - 2

model = SRRegressor(
    binary_operators=[+, ^],
    unary_operators=[exp, cos],
    niterations=40
)
mach = machine(model, X, y)
fit!(mach)

r = report(mach)
r.equations[r.best_idx]

output:

((0.7841441981843509 ^ (x₁ + -3.7374866510975324)) + (exp(x₁ + 27.319351179464586) ^ 0.12208411287569956))

from pysr.

The problem arises because in X negative numbers may occur and, in such case, pow(negate, Real) gives complex number, but the SR is running in Real mode.

from pysr.

Ah yeah that looks like the issue. It simply can't find exactly 2.0000, because even 1.9999 would produce imaginary output, thus creating NaN output and infinite loss – so it would have to randomly guess 2.0000 by pure luck. There are a bunch of solutions:

1. Use * instead of ^.
2. Give the model abs as a unary operator, so you make abs(x)^1.9999 a valid outcome, making it easier to find x^2.
3. Define a custom operator abs_pow(x, y) = abs(x) ^ y (if in Python: in binary_operators as a string, and also with extra_sympy_mappings={"abs_pow": lambda x, y: abs(x) ** y}.
4. Cast the input to imaginary numbers (e.g., + 0j in python or + 0im in Julia), so that -1 ^ 1.5 is a valid result, and so the model can descend smoothly towards -1 ^ 2.
5. I'm assuming this isn't what you want but you could just use square which is one of the unary operators

from pysr.