Is there a way to do some pre-process/post-process every time I call a function in python

I have a number of calls I need to make to functions imported from a package (the package is a shared object file). However I need to do some pre-process / post – process step every time I make a call to a function from this package. Something like this:

import xyz

prepare()
xyz.foo(<args>)
done()

prepare()
xyz.bar(<args>)
done()

prepare()
xyz.foobar()
done()

Is there some way I can ask python to always invoke prepare() before I call a function from xyz module. And also to invoke done() after the call is complete?
Writing prepare and done through my whole code seems redundant and messy. Appreciate your help!

>Solution :

This is typically done with a context manager.

import contextlib

@contextlib.contextmanager
def with_preparation():
    prepare()
    yield
    done()

with preparation():
    xyz.foo(<args>)

with preparation():
    xyz.bar(<args>)

with preparation():
    xyz.foobar()

preparation defines a function that returns a context manager. The with statement works by invoking the context manager’s __enter__ method, then executing the body, then ensuring that the context manager’s __exit__ method is invoked before moving on (whether due to an exception being raised or the body completing normally).

contextlib.contextmanager provides a simple way to define a context manager using a generator function, rather than making you define a class with explicit __enter__ and __exit__ methods.