CODEWITHSUNDEEP
sympy
David Armendariz
David Armendariz

Reputation: 1769

Type of undefined functions in Sympy

Suppose I have the following:

x=Symbol('x')
f=Function('f')(x)
type(f)
f

Why is type(f)=f?

Why it is not UndefinedFunction?

How can I identify this kind of functions more generally?

Upvotes: 2

Views: 772

Answers (1)

user6655984
user6655984

Reputation:

Short answer

It's an instance of AppliedUndef, because it's been "applied" (to variable x). Before that, it was an UndefinedFunction.

Explanation

Let's consider two steps separately: creating a function and applying it. When they are mixed together, we lose the difference between functions and expressions, which is a common source of confusion in SymPy and in math generally.

Unapplied functions

>>> f = Function('f')
>>> type(f)
<class 'sympy.core.function.UndefinedFunction'>

This is an undefined function. Notably, it's a class and not an object. Perhaps it should be an object (a long-standing issue) but on the other hand, this behavior is consistent with defined functions (sin, log) being classes. You can test for f being an UndefinedFunction:

>>> from sympy.core.function import UndefinedFunction
>>> isinstance(f, UndefinedFunction)
True

Applied functions

When you apply f to x, creating f(x), the result is a SymPy expression of class f, just as sin(x) is an expression of class sin. How can we test this expression for being an undefined function applied to something? let's see:

>>> type(f(x)).__mro__
(f, AppliedUndef, Function, Application, <class 'sympy.core.expr.Expr'>, <class 'sympy.core.basic.Basic'>, <class 'sympy.core.evalf.EvalfMixin'>, <class 'object'>)

So, AppliedUndef is the class you want to check here.

>>> from sympy.core.function import AppliedUndef
>>> isinstance(f(x), AppliedUndef)
True

Conclusion

Test with isinstance(..., ...), not with type(...) == ...

Upvotes: 1

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