How to get the caller class name inside a function of another class in python?

Question

My objective is to stimulate a sequence diagram of an application for this I need the information about a caller and callee class names at runtime. I can successfully retrie

Answer 1

Perhaps this is breaking some Python programming protocol, but if Bad is always going to check the class of the caller, why not pass the caller's __class__ to it as part of the call?

class A:

    def Apple(self):
        print "Hello"
        b=B()
        b.Bad(self.__class__)



class B:

    def Bad(self, cls):
        print "dude"
        print "Calling class:", cls


a=A()
a.Apple()

Result:

Hello
dude
Calling class: __main__.A

If this is bad form, and using inspect truly is the preferred way to get the caller's class, please explain why. I'm still learning about deeper Python concepts.

Answer 2

Instead of indexing the return value of inspect.stack(), one could use the method inspect.currentframe(), which avoids the indexing.

prev_frame = inspect.currentframe().f_back
the_class = prev_frame.f_locals["self"].__class__
the_method = prev_frame.f_code.co_name

Answer 3

Python 3.8

import inspect


class B:
    def __init__(self):
        if (parent := inspect.stack()[1][0].f_locals.get('self', None)) and isinstance(parent, A):
            parent.print_coin()


class A:
    def __init__(self, coin):
        self.coin: str = coin
        B()

    def print_coin(self):
        print(f'Coin name: {self.coin}')


A('Bitcoin')

Answer 4

Using the answer from Python: How to retrieve class information from a 'frame' object?

I get something like this...

import inspect

def get_class_from_frame(fr):
  args, _, _, value_dict = inspect.getargvalues(fr)
  # we check the first parameter for the frame function is
  # named 'self'
  if len(args) and args[0] == 'self':
    # in that case, 'self' will be referenced in value_dict
    instance = value_dict.get('self', None)
    if instance:
      # return its class
      return getattr(instance, '__class__', None)
  # return None otherwise
  return None


class A(object):

    def Apple(self):
        print "Hello"
        b=B()
        b.Bad()

class B(object):

    def Bad(self):
        print"dude"
        frame = inspect.stack()[1][0]
        print get_class_from_frame(frame)


a=A()
a.Apple()

which gives me the following output:

Hello
dude
<class '__main__.A'>

clearly this returns a reference to the class itself. If you want the name of the class, you can get that from the __name__ attribute.

Unfortunately, this won't work for class or static methods ...

Answer 5

Well, after some digging at the prompt, here's what I get:

stack = inspect.stack()
the_class = stack[1][0].f_locals["self"].__class__.__name__
the_method = stack[1][0].f_code.co_name

print("I was called by {}.{}()".format(the_class, the_method))
# => I was called by A.a()

When invoked:

➤ python test.py
A.a()
B.b()
  I was called by A.a()

given the file test.py:

import inspect

class A:
  def a(self):
    print("A.a()")
    B().b()

class B:
  def b(self):
    print("B.b()")
    stack = inspect.stack()
    the_class = stack[1][0].f_locals["self"].__class__.__name__
    the_method = stack[1][0].f_code.co_name
    print("  I was called by {}.{}()".format(the_class, the_method))

A().a()

Not sure how it will behave when called from something other than an object.

Answer 6

To store class instance name from the stack to class variable:

import inspect

class myClass():

    caller = ""

    def __init__(self):
        s = str(inspect.stack()[1][4]).split()[0][2:]
        self.caller = s

    def getInstanceName(self):
        return self.caller

This

myClassInstance1 = myClass()
print(myClassInstance1.getInstanceName())

will print:

myClassInstance1