*args and **kwargs
Remarks#
There a few things to note:
-
The names
args
andkwargs
are used by convention, they are not a part of the language specification. Thus, these are equivalent:def func(*args, **kwargs): print(args) print(kwargs)
def func(*a, **b): print(a) print(b)
-
You may not have more than one
args
or more than onekwargs
parameters (however they are not required)def func(*args1, *args2): # File "<stdin>", line 1 # def test(*args1, *args2): # ^ # SyntaxError: invalid syntax
def test(**kwargs1, **kwargs2): # File "<stdin>", line 1 # def test(**kwargs1, **kwargs2): # ^ # SyntaxError: invalid syntax
-
If any positional argument follow
*args
, they are keyword-only arguments that can only be passed by name. A single star may be used instead of*args
to force values to be keyword arguments without providing a variadic parameter list. Keyword-only parameter lists are only available in Python 3.def func(a, b, *args, x, y): print(a, b, args, x, y) func(1, 2, 3, 4, x=5, y=6) #>>> 1, 2, (3, 4), 5, 6
def func(a, b, *, x, y): print(a, b, x, y) func(1, 2, x=5, y=6) #>>> 1, 2, 5, 6
-
**kwargs
must come last in the parameter list.def test(**kwargs, *args): # File "<stdin>", line 1 # def test(**kwargs, *args): # ^ # SyntaxError: invalid syntax
Using *args when writing functions
You can use the star * when writing a function to collect all positional (ie. unnamed) arguments in a tuple:
def print_args(farg, *args):
print("formal arg: %s" % farg)
for arg in args:
print("another positional arg: %s" % arg)
Calling method:
print_args(1, "two", 3)
In that call, farg will be assigned as always, and the two others will be fed into the args tuple, in the order they were received.
Using **kwargs when writing functions
You can define a function that takes an arbitrary number of keyword (named) arguments by using the double star **
before a parameter name:
def print_kwargs(**kwargs):
print(kwargs)
When calling the method, Python will construct a dictionary of all keyword arguments and make it available in the function body:
print_kwargs(a="two", b=3)
# prints: "{a: "two", b=3}"
Note that the **kwargs parameter in the function definition must always be the last parameter, and it will only match the arguments that were passed in after the previous ones.
def example(a, **kw):
print kw
example(a=2, b=3, c=4) # => {'b': 3, 'c': 4}
Inside the function body, kwargs
is manipulated in the same way as a dictionary; in order to access individual elements in kwargs
you just loop through them as you would with a normal dictionary:
def print_kwargs(**kwargs):
for key in kwargs:
print("key = {0}, value = {1}".format(key, kwargs[key]))
Now, calling print_kwargs(a="two", b=1)
shows the following output:
print_kwargs(a = "two", b = 1)
key = a, value = "two"
key = b, value = 1
Using *args when calling functions
A common use case for *args
in a function definition is to delegate processing to either a wrapped or inherited function. A typical example might be in a class’s __init__
method
class A(object):
def __init__(self, b, c):
self.y = b
self.z = c
class B(A):
def __init__(self, a, *args, **kwargs):
super(B, self).__init__(*args, **kwargs)
self.x = a
Here, the a
parameter is processed by the child class after all other arguments (positional and keyword) are passed onto - and processed by - the base class.
For instance:
b = B(1, 2, 3)
b.x # 1
b.y # 2
b.z # 3
What happens here is the class B
__init__
function sees the arguments 1, 2, 3
. It knows it needs to take one positional argument (a
), so it grabs the first argument passed in (1
), so in the scope of the function a == 1
.
Next, it sees that it needs to take an arbitrary number of positional arguments (*args
) so it takes the rest of the positional arguments passed in (1, 2
) and stuffs them into *args
. Now (in the scope of the function) args == [2, 3]
.
Then, it calls class A
’s __init__
function with *args
. Python sees the *
in front of args and “unpacks” the list into arguments. In this example, when class B
’s __init__
function calls class A
’s __init__
function, it will be passed the arguments 2, 3
(i.e. A(2, 3)
).
Finally, it sets its own x
property to the first positional argument a
, which equals 1
.
Using **kwargs when calling functions
You can use a dictionary to assign values to the function’s parameters; using parameters name as keys in the dictionary and the value of these arguments bound to each key:
def test_func(arg1, arg2, arg3): # Usual function with three arguments
print("arg1: %s" % arg1)
print("arg2: %s" % arg2)
print("arg3: %s" % arg3)
# Note that dictionaries are unordered, so we can switch arg2 and arg3. Only the names matter.
kwargs = {"arg3": 3, "arg2": "two"}
# Bind the first argument (ie. arg1) to 1, and use the kwargs dictionary to bind the others
test_var_args_call(1, **kwargs)
Using *args when calling functions
The effect of using the *
operator on an argument when calling a function is that of unpacking the list or a tuple argument
def print_args(arg1, arg2):
print(str(arg1) + str(arg2))
a = [1,2]
b = tuple([3,4])
print_args(*a)
# 12
print_args(*b)
# 34
Note that the length of the starred argument need to be equal to the number of the function’s arguments.
A common python idiom is to use the unpacking operator *
with the zip
function to reverse its effects:
a = [1,3,5,7,9]
b = [2,4,6,8,10]
zipped = zip(a,b)
# [(1,2), (3,4), (5,6), (7,8), (9,10)]
zip(*zipped)
# (1,3,5,7,9), (2,4,6,8,10)
Keyword-only and Keyword-required arguments
Python 3 allows you to define function arguments which can only be assigned by keyword, even without default values. This is done by using star * to consume additional positional parameters without setting the keyword parameters. All arguments after the * are keyword-only (i.e. non-positional) arguments. Note that if keyword-only arguments aren’t given a default, they are still required when calling the function.
def print_args(arg1, *args, keyword_required, keyword_only=True):
print("first positional arg: {}".format(arg1))
for arg in args:
print("another positional arg: {}".format(arg))
print("keyword_required value: {}".format(keyword_required))
print("keyword_only value: {}".format(keyword_only))
print(1, 2, 3, 4) # TypeError: print_args() missing 1 required keyword-only argument: 'keyword_required'
print(1, 2, 3, keyword_required=4)
# first positional arg: 1
# another positional arg: 2
# another positional arg: 3
# keyword_required value: 4
# keyword_only value: True
Populating kwarg values with a dictionary
def foobar(foo=None, bar=None):
return "{}{}".format(foo, bar)
values = {"foo": "foo", "bar": "bar"}
foobar(**values) # "foobar"
**kwargs and default values
To use default values with **kwargs
def fun(**kwargs):
print kwargs.get('value', 0)
fun()
# print 0
fun(value=1)
# print 1