112

Is it possible to have overloaded functions in Python?

In C# I would do something like:

void myfunction (int first, string second)
{
    // Some code
}

void myfunction (int first, string second, float third)
{
    // Some different code
}

And then when I call the function it would differentiate between the two based on the number of arguments. Is it possible to do something similar in Python?

4

6 Answers 6

121

EDIT For the new single dispatch generic functions in Python 3.4, see http://www.python.org/dev/peps/pep-0443/

You generally don't need to overload functions in Python. Python is dynamically typed, and supports optional arguments to functions.

def myfunction(first, second, third = None):
    if third is None:
        #just use first and second
    else:
        #use all three

myfunction(1, 2) # third will be None, so enter the 'if' clause
myfunction(3, 4, 5) # third isn't None, it's 5, so enter the 'else' clause
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14 Comments

But note that calling different functions based on the type of the arguments is much more difficult (although not impossible).
Well, there is a difference between overloading/polymorphism and conditionals. Do you mean that we do not need the polymorphysm since we have the conditionals?
@Val I'm saying, basically, that in a dynamically typed language you don't need overloading as a language feature because you can trivially emulate it in code, and so can get polymorphism that way.
@Val I'm saying that between optional arguments and dynamic types the way you have them in Python, you don't need Java style method overloading to achieve polymorphism.
@navidoo I would argue that isn't a good idea -- a function shoudn't return completely different types of things. However, you can always define a local generator function inside the main function, call it, and return the resulting generator -- from the outside it will be the same as if the main function was a generator. Example here.
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62

In normal Python you can't do what you want. There are two close approximations:

def myfunction(first, second, *args):
    # 'args' is a tuple of extra arguments

def myfunction(first, second, third=None):
    # 'third' is optional

However, if you really want to do this, you can certainly make it work (at the risk of offending the traditionalists ;o). In short, you would write a wrapper(*args) function that checks the number of arguments and delegates as appropriate. This kind of "hack" is usually done via decorators. In this case, you could achieve something like:

@overload
def myfunction(first):
    ....

@myfunction.overload
def myfunction(first, second):
    ....

@myfunction.overload
def myfunction(first, second, third):
    ....

And you'd implement this by making the overload(first_fn) function (or constructor) return a callable object where the __call__(*args) method does the delegation explained above and the overload(another_fn) method adds extra functions that can be delegated to.

You can see an example of something similar here http://acooke.org/pytyp/pytyp.spec.dispatch.html, but that is overloading methods by type. It's a very similar approach...

And something similar (using argument types) is being added to Python 3 - PEP 443 -- Single-dispatch generic functions

4 Comments

The range builtin uses overload, so is it really a hack? github.com/python/typeshed/blob/master/stdlib/2and3/…
Is this a mistake: @myfunction.overload? Should not just all of them have @overload, and you would then need another function without the @overload on top of this? I have just tested @overload in quite a few settings. I wonder whether this answer is outdated. typing does not allow you to overload in the code, it is just to show what the type hints would give you if you overloaded the function. Which you do not, since typing is nothing about the working code, instead, it is about making code quickly readable.
I dared to answer at How can I type-hint a function where the return type depends on the input type of an argument?. Correct me if I am wrong, but typing, in my humble opinion, does not change the working of the code, it is just to quickly show what is going on. There is no overloading with typing, the typing overload functions are not run at all, as far as I can see. I could be wrong.
Whoever approved the edit that added from typing import overload to this answer - you are WRONG. This answer does not talk about typing.overload, but rather about a hypothetical overload function one could write. It is called functools.singledispatch in modern python, but this answer predates addition of that helper. It has nothing to do with typing.overload, not to mention that typing.overload is even younger than functools.singledispatch. Please don't do that again. The added import makes this reasonable answer completely broken and misleading.
11

Yes, it's possible. I wrote the code below in Python 3.2.1:

def overload(*functions):
    return lambda *args, **kwargs: functions[len(args)](*args, **kwargs)

Usage:

myfunction=overload(no_arg_func, one_arg_func, two_arg_func)

Note that the lambda returned by the overload functions choose a function to call depending on the number of unnamed arguments.

The solution isn't perfect, but at the moment I can't write anything better.

1 Comment

Great way to refresh the args and kwargs thingy... Yet, is it pythonic to make it to everyday code? I doubt it. I guess in Python, they prefer try and except or if and else or optional (=None) over this small overload overhead, so that your good code was avoided from the start of Python.
2

Here is the way to overload python functions with default arguments as well as keyword arguments

from multipledispatch import dispatch

# FOR hi(a: int, b: int = 3)

@dispatch(int, int)
def _hi(a: int, b: int):
    print(a, b)


@dispatch(int, int)
def hi(a: int, b: int = 3):
    _hi(a, b)


@dispatch(int, b=int)
def hi(a: int, *, b: int = 3):
    _hi(a, b)


# FOR hi(a: str, b: int = 3)

@dispatch(str, int)
def _hi(a: str, b: int):
    print(a, b, 'str!')


@dispatch(str, int)
def hi(a: str, b: int = 3):
    _hi(a, b)


@dispatch(str, b=int)
def hi(a: str, *, b: int = 3):
    _hi(a, b)

hi(2)
hi(2, 3)
hi(2, b=3)
hi('2')
hi('2', 3)
hi('2', b=3)

Output

2 3
2 3
2 3
2 3 str!
2 3 str!
2 3 str!

Comments

1

It is not possible directly. You can use explicit type checks on the arguments given though, although this is generally frowned upon.

Python is dynamic. If you are unsure what an object can do, just try: and call a method on it, then except: errors.

If you don't need to overload based on types, but just on the number of arguments, use keyword arguments.

1 Comment

I don't think he knew about optional arguments.
0

Overloading methods is tricky in Python. However, there could be usage of passing the dict, list or primitive variables.

I have tried something for my use cases, and this could help here to understand people to overload the methods.

Let's take the example use in one of the Stack Overflow questions:

A class overload method with call the methods from different class.

def add_bullet(sprite=None, start=None, headto=None, spead=None, acceleration=None):

Pass the arguments from a remote class:

add_bullet(sprite = 'test', start=Yes, headto={'lat':10.6666, 'long':10.6666}, accelaration=10.6}

Or

add_bullet(sprite = 'test', start=Yes, headto={'lat':10.6666, 'long':10.6666}, speed=['10','20,'30']}

So, handling is being achieved for list, Dictionary or primitive variables from method overloading.

Try it out for your code.

1 Comment

Probably not accelaration. More like acceleration (an ae).

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