Confusion with custon operator definition

[anubhavkamal]

def p(i):
if i < 20.0:
return sympy.log(27.1285645162762 - 1.42277605519119i)
else:
return float("nan")
model = pysr.PySRRegressor(
turbo=True,
maxsize=30,
batching=True,
warm_start=True,
model_selection='best',
niterations=40000,
precision = 64,
binary_operators=['+','-','','/'],
#unary_operators=['exp', 'log'],
unary_operators = ['exp', "fn(x) = log(27.1285645162762 - 1.42277605519119*x)"],
extra_sympy_mappings = {"fn": lambda x: p(x)},
nested_constraints = {"exp": {"exp" : 0}},
#loss = 'LogitDistLoss()',
)

   Hello!
   I have defined a custom function of the form shown above but PySR is throwing following error - 
   
   RuntimeError: <PyCall.jlwrap (in a Julia function called from Python)

JULIA: The operator fn is not well-defined over the real line, as it threw the error DomainError when evaluating the input 20.408163265306122. You can work around this by returning NaN for invalid inputs. For example, safe_log(x::T) where {T} = x > 0 ? log(x) : T(NaN).

The inputs in the dataset is less than 18. Thus, there is no case where the value of log(x) blows to infinity. I am not sure about this 'DomainError'.

[MilesCranmer]

The inputs in the dataset is less than 18. Thus, there is no case where the value of log(x) blows to infinity. I am not sure about this 'DomainError'.

Keep in mind it’s going to try many different things as input to your operator; not just the dataset feature itself. Because you have / it could also scale your features. So you always want to catch values before they cause a domain error and return NaN instead.

cheers,
Miles

[anubhavkamal]

Hello!
Thanks for the swift reply.

While defining other "questionable" functions such as 'log(x)', '1/log(x)', '1/x'. I didn't face this domainError.
What shall I change here, if I want a particular structure in my function?
Thanks!

[MilesCranmer]

Some functions get automatically mapped to "safe" versions. Like if you write log as one of the operators, it will get automatically converted to x > 0 ? log(x) : NaN. But for custom functions this conversion doesn't happen.

[anubhavkamal]

Thanks @MilesCranmer !
Turns out to be an easier fix. Just used 'absolute' value function before passing it to the log function.
Your answers helped!