*args and **kwargs in Python (Variable-length arguments) | note.nkmk.me (2024)

In Python, you can specify a variable number of arguments when calling a function by prefixing the parameter names with * or ** in the function definition.

By convention, *args (arguments) and **kwargs (keyword arguments) are often used as parameter names, but you can use any name as long as it is prefixed with * or **. The sample code in this article uses *args and **kwargs.

See the following article for the basics of functions in Python.

• Define and call functions in Python (def, return)

Additionally, using * and ** when calling a function allows you to unpack and pass lists and dictionaries as arguments.

• Unpack and pass list, tuple, dict to function arguments in Python

*args: Receive multiple arguments as a tuple

By prefixing a parameter with *, such as *args, in the function definition, a function can accept any number of arguments.

def my_sum(*args): return sum(args)print(my_sum(1, 2, 3, 4))# 10print(my_sum(1, 2, 3, 4, 5, 6, 7, 8))# 36

Within the function, multiple arguments are received as a tuple. In this example, the tuple is then passed to the sum() function to calculate the sum.

def my_sum2(*args): print('args: ', args) print('type: ', type(args)) print('sum : ', sum(args))my_sum2(1, 2, 3, 4)# args: (1, 2, 3, 4)# type: <class 'tuple'># sum : 10

*args can be used with positional arguments.

The values specified after the positional arguments are passed as a tuple to args. If no extra positional arguments are passed, args will be an empty tuple.

def func_args(arg1, arg2, *args): print('arg1: ', arg1) print('arg2: ', arg2) print('args: ', args)func_args(0, 1, 2, 3, 4)# arg1: 0# arg2: 1# args: (2, 3, 4)func_args(0, 1)# arg1: 0# arg2: 1# args: ()

You can place *args anywhere in the parameter list. However, any parameters defined after *args must be specified using the keyword format parameter_name=value when calling the function.

For example, in the following sample code, the last value passed to the function (in this case, 4) must be specified as a keyword argument (arg2=4). If it is not, a TypeError will be raised.

def func_args2(arg1, *args, arg2): print('arg1: ', arg1) print('arg2: ', arg2) print('args: ', args)# func_args2(0, 1, 2, 3, 4)# TypeError: func_args2() missing 1 required keyword-only argument: 'arg2'func_args2(0, 1, 2, 3, arg2=4)# arg1: 0# arg2: 4# args: (1, 2, 3)

With this property, you can define a parameter named * and treat any subsequent parameters as keyword-only arguments.

def func_args_kw_only(arg1, *, arg2): print('arg1: ', arg1) print('arg2: ', arg2)# func_args_kw_only(100, 200)# TypeError: func_args_kw_only() takes 1 positional argument but 2 were givenfunc_args_kw_only(100, arg2=200)# arg1: 100# arg2: 200

**kwargs: Receive multiple keyword arguments as a dictionary

By prefixing a parameter with **, such as **kwargs, in the function definition, a function can accept any number of keyword arguments.

Within the function, multiple keyword arguments are received as a dictionary, with the argument names as keys and their values as the corresponding dictionary values.

def func_kwargs(**kwargs): print('kwargs: ', kwargs) print('type: ', type(kwargs))func_kwargs(key1=1, key2=2, key3=3)# kwargs: {'key1': 1, 'key2': 2, 'key3': 3}# type: <class 'dict'>

**kwargs can also be used alongside positional arguments.

def func_kwargs_positional(arg1, arg2, **kwargs): print('arg1: ', arg1) print('arg2: ', arg2) print('kwargs: ', kwargs)func_kwargs_positional(0, 1, key1=1)# arg1: 0# arg2: 1# kwargs: {'key1': 1}

When calling a function, prefixing a dictionary with ** allows you to pass its elements as individual arguments.

d = {'key1': 1, 'key2': 2, 'arg1': 100, 'arg2': 200}func_kwargs_positional(**d)# arg1: 100# arg2: 200# kwargs: {'key1': 1, 'key2': 2}

See the following article for details on unpacking function arguments.

• Unpack and pass list, tuple, dict to function arguments in Python

A parameter prefixed with ** can only be defined at the end of the parameter list. If you define other parameters after the parameter, a SyntaxError will be raised.

# def func_kwargs_error(**kwargs, arg):# print(kwargs)# SyntaxError: invalid syntax