TypeError: method() takes 1 positional argument but 2 were given

Question

If I have a class...

class MyClass:

    def method(arg):
        print(arg)

...which I use to create an object...

my_objec         


        

Answer 1

In my case, I forgot to add the ()

I was calling the method like this

obj = className.myMethod

But it should be is like this

obj = className.myMethod()

Answer 2

It occurs when you don't specify the no of parameters the __init__() or any other method looking for.

For example:

class Dog:
    def __init__(self):
        print("IN INIT METHOD")

    def __unicode__(self,):
        print("IN UNICODE METHOD")

    def __str__(self):
        print("IN STR METHOD")

obj=Dog("JIMMY",1,2,3,"WOOF")

When you run the above programme, it gives you an error like that:

TypeError: __init__() takes 1 positional argument but 6 were given

How we can get rid of this thing?

Just pass the parameters, what __init__() method looking for

class Dog:
    def __init__(self, dogname, dob_d, dob_m, dob_y, dogSpeakText):
        self.name_of_dog = dogname
        self.date_of_birth = dob_d
        self.month_of_birth = dob_m
        self.year_of_birth = dob_y
        self.sound_it_make = dogSpeakText

    def __unicode__(self, ):
        print("IN UNICODE METHOD")

    def __str__(self):
        print("IN STR METHOD")


obj = Dog("JIMMY", 1, 2, 3, "WOOF")
print(id(obj))

Answer 3

In Python, this:

my_object.method("foo")

...is syntactic sugar, which the interpreter translates behind the scenes into:

MyClass.method(my_object, "foo")

...which, as you can see, does indeed have two arguments - it's just that the first one is implicit, from the point of view of the caller.

This is because most methods do some work with the object they're called on, so there needs to be some way for that object to be referred to inside the method. By convention, this first argument is called self inside the method definition:

class MyNewClass:

    def method(self, arg):
        print(self)
        print(arg)

If you call method("foo") on an instance of MyNewClass, it works as expected:

>>> my_new_object = MyNewClass()
>>> my_new_object.method("foo")
<__main__.MyNewClass object at 0x29045d0>
foo

Occasionally (but not often), you really don't care about the object that your method is bound to, and in that circumstance, you can decorate the method with the builtin staticmethod() function to say so:

class MyOtherClass:

    @staticmethod
    def method(arg):
        print(arg)

...in which case you don't need to add a self argument to the method definition, and it still works:

>>> my_other_object = MyOtherClass()
>>> my_other_object.method("foo")
foo

Answer 4

Pass cls parameter into @classmethod to resolve this problem.

@classmethod
def test(cls):
    return ''

Answer 5

Something else to consider when this type of error is encountered:

I was running into this error message and found this post helpful. Turns out in my case I had overridden an __init__() where there was object inheritance.

The inherited example is rather long, so I'll skip to a more simple example that doesn't use inheritance:

class MyBadInitClass:
    def ___init__(self, name):
        self.name = name

    def name_foo(self, arg):
        print(self)
        print(arg)
        print("My name is", self.name)


class MyNewClass:
    def new_foo(self, arg):
        print(self)
        print(arg)


my_new_object = MyNewClass()
my_new_object.new_foo("NewFoo")
my_bad_init_object = MyBadInitClass(name="Test Name")
my_bad_init_object.name_foo("name foo")

Result is:

<__main__.MyNewClass object at 0x033C48D0>
NewFoo
Traceback (most recent call last):
  File "C:/Users/Orange/PycharmProjects/Chapter9/bad_init_example.py", line 41, in <module>
    my_bad_init_object = MyBadInitClass(name="Test Name")
TypeError: object() takes no parameters

PyCharm didn't catch this typo. Nor did Notepad++ (other editors/IDE's might).

Granted, this is a "takes no parameters" TypeError, it isn't much different than "got two" when expecting one, in terms of object initialization in Python.

Addressing the topic: An overloading initializer will be used if syntactically correct, but if not it will be ignored and the built-in used instead. The object won't expect/handle this and the error is thrown.

In the case of the sytax error: The fix is simple, just edit the custom init statement:

def __init__(self, name):
    self.name = name

Answer 6

Newcomer to Python, I had this issue when I was using the Python's ** feature in a wrong way. Trying to call this definition from somewhere:

def create_properties_frame(self, parent, **kwargs):

using a call without a double star was causing the problem:

self.create_properties_frame(frame, kw_gsp)

TypeError: create_properties_frame() takes 2 positional arguments but 3 were given

The solution is to add ** to the argument:

self.create_properties_frame(frame, **kw_gsp)