Hence, we conclude that Python Function Arguments and its three types of arguments to functions. These are- default, keyword, and arbitrary arguments.
Python Default Arguments Here is an example. def greet(name, msg="Good morning!"): """ This function greets to the person with the provided message. If the message is not provided, it defaults to "Good morning!" """ print("Hello", name + ', ' + msg) greet("Kate") greet("Bruce", "How do you do?")
5 Types of Arguments in Python Function Definition:positional arguments. arbitrary positional arguments. arbitrary keyword arguments.
A default argument is a value provided in a function declaration that is automatically assigned by the compiler if the calling function doesn't provide a value for the argument. In case any value is passed, the default value is overridden.
Just put the default arguments before the *args
:
def foo(a, b=3, *args, **kwargs):
Now, b
will be explicitly set if you pass it as a keyword argument or the second positional argument.
Examples:
foo(x) # a=x, b=3, args=(), kwargs={}
foo(x, y) # a=x, b=y, args=(), kwargs={}
foo(x, b=y) # a=x, b=y, args=(), kwargs={}
foo(x, y, z, k) # a=x, b=y, args=(z, k), kwargs={}
foo(x, c=y, d=k) # a=x, b=3, args=(), kwargs={'c': y, 'd': k}
foo(x, c=y, b=z, d=k) # a=x, b=z, args=(), kwargs={'c': y, 'd': k}
Note that, in particular, foo(x, y, b=z)
doesn't work because b
is assigned by position in that case.
This code works in Python 3 too. Putting the default arg after *args
in Python 3 makes it a "keyword-only" argument that can only be specified by name, not by position. If you want a keyword-only argument in Python 2, you can use @mgilson's solution.
The syntax in the other question is python3.x only and specifies keyword only arguments. It doesn't work on python2.x.
For python2.x, I would pop
it out of kwargs:
def func(arg1, arg2, *args, **kwargs):
opt_arg = kwargs.pop('opt_arg', 'def_val')
Similar approach to @yaccob, but clear and concise:
In Python 3.5 or greater:
def foo(a, b=3, *args, **kwargs):
defaultKwargs = { 'c': 10, 'd': 12 }
kwargs = { **defaultKwargs, **kwargs }
print(a, b, args, kwargs)
# Do something
foo(1) # 1 3 () {'c': 10, 'd': 12}
foo(1, d=5) # 1 3 () {'c': 10, 'd': 5}
foo(1, 2, 4, d=5) # 1 2 (4,) {'c': 10, 'd': 5}
Note: you can use In Python 2
kwargs = merge_two_dicts(defaultKwargs, kwargs)
In Python 3.5
kwargs = { **defaultKwargs, **kwargs }
In Python 3.9
kwargs = defaultKwargs | kwargs # NOTE: 3.9+ ONLY
You could also use a decorator like this:
import functools
def default_kwargs(**defaultKwargs):
def actual_decorator(fn):
@functools.wraps(fn)
def g(*args, **kwargs):
defaultKwargs.update(kwargs)
return fn(*args, **defaultKwargs)
return g
return actual_decorator
Then just do:
@default_kwargs(defaultVar1 = defaultValue 1, ...)
def foo(*args, **kwargs):
# Anything in here
For instance:
@default_kwargs(a=1)
def f(*args, **kwargs):
print(kwargs['a']+ 1)
f() # Returns 2
f(3) # Returns 4
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