Adding lambda functions to a symbol table of a class [duplicate]

Question

I am trying to add a lambda function to the symbol table of a class:

class B:
    def __init__(self):
        func = lambda self, x: x*x  # <-- Adding 'self' produces error
        self.__dict__['test1'] = func
        print(self.test1(2))
        print(self.test2(2))

    def test2(self, b):
        return b*b*b

b = B()

but this produces an error (running the script with python t.py):

Traceback (most recent call last):
  File "./t.py", line 14, in <module>
    b = B()
  File "./t.py", line 8, in __init__
    print(self.test1(2))
TypeError: <lambda>() missing 1 required positional argument: 'x'

However, if I remove self as an argument to the Lambda function, it works fine.

Why isn't self required as an argument for the Lambda function here?

You may use self.test1 = types.MethodType(func, self), otherwise it will acts like a static method, without Python applying the bit of magic necessary for self to refer to the class instance. — Delgan
– Delgan, Commented Jun 24, 2017 at 18:00

Ilario Pierbattista · Accepted Answer · 2017-10-30 22:46:35Z

1

Class methods' calls:

instance.method(arg1, arg2, ..., argn)

are internally binded to:

instance_class.method(instance, arg1, ..., argn)

where, obviously, instance is self.

You are storing your lambda as an instance attribute, which wouldn't be treated like an instance method. That's why self is not passed automatically as first argument.

edited Oct 30, 2017 at 22:46

answered Jun 24, 2017 at 18:05

Ilario Pierbattista

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3 Comments

Håkon Hægland Over a year ago

Thanks for the answer. This makes sense. So adding the lambda function as an instance attribute does not make it an instance method? But how could one make the lambda be treated as an instance method?

Ilario Pierbattista Over a year ago

The answer to this question explains it very well. Try self.__dict__['test1'] = types.MethodType( func, self )

Håkon Hægland Over a year ago

Thanks @IlarioPierbattista. I think I found a solution: according to the answer you referred to, I should be able to make it an instance method using: self.__dict__['test1'] = types.MethodType( func, self )

melpomene · Accepted Answer · 2017-06-24 18:07:29Z

1

lambda functions are anonymous functions, they are not bound to the class, so in this case you are creating a lambda with 2 arguments (self and x). The error is clearly telling you that when you call the lambda function you are just passing one argument bound to self, so you lack the x.

edited Jun 24, 2017 at 18:07

melpomene

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answered Jun 24, 2017 at 18:06

Netwave

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