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 retrieve the caller function but not able to get a caller class name?

#Scenario caller.py:

import inspect

class A:

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



class B:

    def Bad(self):
        print"dude"
        print inspect.stack()


a=A()
a.Apple()

When I printed the stack there was no information about the caller class. So is it possible to retrieve the caller class during runtime ?

What do you mean by "caller class". Do you mean `a`, `A`, `b` or `B`? — mgilson, Jun 12 '13 at 12:18
@mgilson What i meant is when the code is running in the method "def Bad" which is under the class B (callee) i must be able to print retrieve the name of the caller class which is "A" in this case. — Kaushik, Jun 12 '13 at 12:21
@mgilson I can print "inspect.stack()[1][3]" statement which would get me only the caller function. — Kaushik, Jun 12 '13 at 12:22
Similar, related, interesting to see: [get a class name of calling method](https://stackoverflow.com/questions/53153075/get-a-class-name-of-calling-method/53153512) — artu-hnrq, Dec 17 '21 at 03:43

score 63 · Accepted Answer · edited Jan 14 '20 at 17:47

63

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.

edited Jan 14 '20 at 17:47

M. Gruber

351
3
10

answered Jun 12 '13 at 12:33

brice

24,329
7
79
95

1

Nicely done. At some point I'd like to become more familiar with code and frame objects... – mgilson Jun 12 '13 at 12:43
4

Note that this won't work for a static or classmethod either as `self` then won't be defined, or it won't be an instance of the class (most likely) – mgilson Jun 12 '13 at 12:47
@mgilson: can you suggest a better way?' – Swadhikar Mar 05 '17 at 07:07
2

This work of the method is invoked via `super` in a subclass. Or at least it won't be what you expect since `type(self)` will be the type of the subclass, not the the type of the class which actually owns the method. – rmorshea Jun 10 '17 at 20:53
stack[1][0].f_locals["self"].__class__ returns type instead of str. For getting class name in string a cast is needed the_class = str(stack[1][0].f_locals["self"].__class__) – Code_Worm Jan 25 '19 at 21:17
2

A more accurate way to get the caller class is stack[1][0].f_locals["__class\_\_"]. The calling class is returned, not the class of the first argument. – Elijah May 20 '20 at 16:59
2

This was extremely useful to me. I just wanted to give a sincere thank-you. – duma Oct 08 '21 at 13:50

score 8 · Answer 2 · edited May 23 '17 at 12:02

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 ...

This method is actually more perfect since it checks for "self" and if not displays "None". This is exactly what I needed. — Kaushik, Jun 12 '13 at 12:59

score 6 · Answer 3 · answered Nov 27 '18 at 00:13

6

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

answered Nov 27 '18 at 00:13

djeongo

61
1
1

1

This would print something like `I was called by .your_method`. To achieve the same result from accepted answer, change the variable `the_class` to `prev_frame.f_locals['self'].__class__.__name__` – igorkf Apr 08 '21 at 22:51

score 5 · Answer 4 · answered Jun 12 '13 at 12:43

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.

score 1 · Answer 5 · answered Apr 07 '20 at 10:12

1

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')

answered Apr 07 '20 at 10:12

Джон Смит

21
6

1

Never thought calling a method or accessing attributes from the caller's class was possible. This provides interesting flexibility. – JWCompDev Feb 19 '22 at 07:18
@jlwise24 You are right, it is very interesting, but sometimes it is slow. You have to be careful, especially in Python. Read about "introspection and reflection" in programming. – Джон Смит Feb 19 '22 at 17:10

score 0 · Answer 6 · answered Apr 30 '18 at 06:32

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

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

6 Answers6

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