This question is about the id method in Python and if a class method is the same instance for the class object itself and any object instances created from that class.

Sample code:

class Person:

    def __init__(self, name: str) -> None:
        self.name = name

    def name_as_upper(self) -> str:
        return self.name.upper()

p = Person("john")
p2 = Person("sarah")
print(
    id(p.name_as_upper), id(p2.name_as_upper), id(Person.name_as_upper)
)  # first two are the same, last one different

print(p.name_as_upper())  # JOHN
print(Person.name_as_upper(p))  # JOHN

Person.name_as_upper = (
    lambda self: self.name.upper() + " - suffix"
)  # "point" name_as_upper method object to different method

print(p.name_as_upper())  # JOHN - suffix
print(Person.name_as_upper(p))  # JOHN - suffix

In the above example, the ids printed by the calls id(p.name_as_upper) and id(p2.name_as_upper) are identical, but the call id(Person.name_as_upper) is different.

I was expecting all of the ids to be identical. The method object referenced by the name Person.name_as_upper is different?

I thought a call to a method on an object instance was translated into a call to the method object belonging to the class, i.e.:

p.name_as_upper()
# Would be translated / equivalent to
Person.name_as_upper(p)

When re-assigning Person.name_as_upper to a new method (the lambda expression in the sample code above) the change is seen also by the instance p.
So p.name_is_upper still is “the same” as Person.name_as_upper.

Thankful for any clarifications to my (lack of) understanding.