Methods with the same name in one class in Python

Question

How can I declare a few methods with the same name, but with different numbers of parameters or different types in one class?

What must I change in the following class?

class MyClass:
    """"""

    #----------------------------------------------------------------------
    def __init__(self):
        """Constructor"""
    def my_method(self,parameter_A_that_Must_Be_String):
        print parameter_A_that_Must_Be_String

    def my_method(self,parameter_A_that_Must_Be_String,parameter_B_that_Must_Be_String):
        print parameter_A_that_Must_Be_String
        print parameter_B_that_Must_Be_String

    def my_method(self,parameter_A_that_Must_Be_String,parameter_A_that_Must_Be_Int):
        print parameter_A_that_Must_Be_String * parameter_A_that_Must_Be_Int

Answer 1

You can have a function that takes in a variable number of arguments.

def my_method(*args, **kwds):
    # Do something

# When you call the method
my_method(a1, a2, k1=a3, k2=a4)

# You get:
args = (a1, a2)
kwds = {'k1':a3, 'k2':a4}

So you can modify your function as follows:

def my_method(*args):
    if len(args) == 1 and isinstance(args[0], str):
        # Case 1
    elif len(args) == 2 and isinstance(args[1], int):
        # Case 2
    elif len(args) == 2 and isinstance(args[1], str):
        # Case 3

Answer 2

You can't. There are not overloads or multimethods or similar things. One name refers to one thing. As far as the language is concerned anyway, you can always emulate them yourself... You could check types with isinstance (but please do it properly - e.g. in Python 2, use basestring to detect both strings and unicode), but it's ugly, generally discouraged and rarely useful. If the methods do different things, give them different names. Consider polymorphism as well.

Answer 3

Using Python 3.5 or higher, you can use @typing.overload to provide type annotations for overloaded functions/methods.

From the docs:

@overload
def process(response: None) -> None:
    ...
@overload
def process(response: int) -> tuple[int, str]:
    ...
@overload
def process(response: bytes) -> str:
    ...
def process(response):
    <actual implementation>

Answer 4

Short answer: you can't (see this previous discussion). Typically you'd use something like (you could add more type checking and reorder):

def my_method(self,parameter_A, parameter_B=None):
  if isinstance(parameter_B, int):
    print parameter_A * parameter_B
  else:
    print parameter_A
    if parameter_B is not None:
      print parameter_B

Answer 5

You can try multimethods in Python:

http://www.artima.com/weblogs/viewpost.jsp?thread=101605

But I don't believe multimethod is a way to go. Rather objects that you pass to a method should have common interface. You are trying to achieve method overloading similar to the one in C++, but it is very rarely required in Python. One way to do this is a cascade of ifs using isinstance, but that's ugly.

Answer 6

Python is nothing like Java.

There are not really types, just objects with methods.

There is a way to test if a passed object is from a class, but it is mainly bad practices.

However, the code you want to produce for the two first methods should be something like

class MyClass(object):
    def my_method(self, str1, str2=None):
        print str1
        if str2: print str2

For the third, well... Use a different name...

Answer 7

This cannot work. No matter how many arguments you have, the name m will be overriden with the second m method.

class C:
    def m(self):
        print('m first')
    def m(self, x):
        print(f'm second {x}')


ci=C();
#ci.m() # will not work TypeError: m() missing 1 required positional argument: 'x'
ci.m(1) # works

---

The output will simple be:

m second 1

Answer 8

As of Python 3.10 a more elegant solution would be to use Structural Pattern Matching.

def my_method(parameters):
    match parameters:
        case str():
            # Case 1
        case (str(), str()):
            # Case 2
        case (str(), int()):
            # Case 3
        case _:
            print('no match')

Answer 9

class MyClass:
    def __init__(this, foo_str, bar_int):
        this.__foo = foo_str
        this.__bar = bar_int

    def foo(this, new=None):
        if new != None:
            try:
                this.__foo = str(new)
            except ValueError:
                print("Illegal value. foo unchanged.")

        return this.__foo

    def bar(this, new=None):
        if new != None:
            try:
                this.__bar = int(new)
            except ValueError:
                print("Illegal value. bar unchanged.")

        return this.__bar

obj = MyClass("test", 42)
print(obj.foo(), obj.bar())

print(obj.foo("tset"), obj.bar(24))

print(obj.foo(42), obj.bar("test"))

Output:
    test 42
    tset 24
    Illegal value. bar unchanged.
    42 24

Answer 10

I think one very simple example is missing from all the answers, and that is: what to do when the only difference between variations on the method is the number of arguments. The answer still is to use a method with variable number of arguments.

Say, you start with a method that requires use of two arguments

def method(int_a, str_b):
    print("Got arguments: '{0}' and '{1}'".format(int_a, str_b)

then you need to add a variant with just the second argument (say, because the integer is redundant), the solution is very simple:

def _method_2_param(int_a, str_b):
    print("Got arguments: '{0}' and '{1}'".format(int_a, str_b))

def _method_1_param(str_b):
    print("Got argument: '{0}'".format(str_b))

def method(*args, **kwargs):
    if len(args) + len(kwargs) == 2:
        return _method_2_param(args, kwargs)
    elif len(args) + len(kwargs) == 1:
        return _method_1_param(args, kwargs)
    else:
        raise TypeError("Method requires one or two arguments")

The nice thing about this solution is that no matter if the calling code used keyword arguments or positional arguments before, it will still continue to work.

Answer 11

You probably want a pattern similar to the following: Note that adding '_' to the beginning of a method name is convention for marking a private method.

class MyClass:
    """"""

    #----------------------------------------------------------------------
    def __init__(self):
        """Constructor"""
    def my_method(self,parameter_A_that_Must_Be_String, param2=None):
        if type(param2) == str:
            return self._my_method_extra_string_version(parameter_A_that_Must_Be_String, param2)
        elif type(param2) == int:
            return self._my_method_extra_int_version(parameter_A_that_Must_Be_String, param2)
        else:
            pass # use the default behavior in this function
        print parameter_A_that_Must_Be_String

    def _my_method_extra_string_version(self,parameter_A_that_Must_Be_String, parameter_B_that_Must_Be_String):
        print parameter_A_that_Must_Be_String
        print parameter_B_that_Must_Be_String

    def _my_method_extra_int_version(self,parameter_A_that_Must_Be_String, parameter_A_that_Must_Be_Int):
        print parameter_A_that_Must_Be_String * parameter_A_that_Must_Be_Int

Answer 12

I think @overload solves the thing: VS Code -> f12 on open() -> it shows a bunch of duplicates of the open() function:

@overload
def open(...):
@overload
def open(...):
...

So, here's the result...

from typing import overload

class MyClass:
""""""

#----------------------------------------------------------------------
def __init__(self):
    """Constructor"""
@overload
def my_method(self,parameter_A_that_Must_Be_String,parameter_B_that_Must_Be_String):
    print parameter_A_that_Must_Be_String
    print parameter_B_that_Must_Be_String
@overload
def my_method(self,parameter_A_that_Must_Be_String,parameter_A_that_Must_Be_Int):
    print parameter_A_that_Must_Be_String * parameter_A_that_Must_Be_Int
def my_method(self,parameter_A_that_Must_Be_String):
    print parameter_A_that_Must_Be_String

I'm sorry for being nervous...