The use of *[parameter name] and **[parameter name] of Python functions

1. *[Parameter name]

transfer

Legal call

Ordinary call

*参数名Generally written *args, such as:

def func(*args):
	print(args)

You can try to call func:

>>> func(1)
(1,)
>>> func()
()
>>> func(1, 2, 3)
(1, 2, 3)
>>> func(dict(), set(), str(), int())
({
    
    }, set(), '', 0)

Therefore, we found that such a function can pass any number of parameters (including 0), and *will pack the parameters into one tuple, such as (1,) () (1, 2, 3) ({}, set(), '', 0).

Special call

What if you already have an tupleobject and want to pass argsit in?
First define an tupleobject:

>>> tuple_object = (1, 2, 3)
>>> print(tuple_object)
(1, 2, 3)

Wrong way

Generally think of this method:

>>> func(tuple_object)
((1, 2, 3),)

((1, 2, 3),)? Shouldn't it be (1, 2, 3)?
Because the system recognizes it as a argsparameter in the middle, it argsis " tuplein the middle tuple", which is wrong. OH NO!

The right way

To tuple_objectplaying in front of a *, OK:

>>> func(*tuple_object)
(1, 2, 3)

This is "unpacking".

Illegal call

What if it is called func(a=1, b=2)? Please see:

>>> func(a=1, b=2)

Get TypeError:

Traceback (most recent call last):
  File "<*args test file>", line 1, in <module>
    func(a=1, b=2)
TypeError: func() got an unexpected keyword argument 'a'

keyword argumentWhat is in the error ?

• keyword argumentIt is a a=1 b=2 c='Hi'form of passing parameters like this.
• Simply put, it keyword argumentis a name=valueform of transfer of parameters.

Therefore, you should use valueformal parameter transfers (in English positional argument) instead of using name=valueparameter transfers.

Default parameters

*[参数名]The parameters cannot have default parameters:

as shown in the figure above, try to set the default parameters will report SyntaxError, if you really want to set the default parameters, you should use a method similar to "manually set default values":

# 手动设置*args的参数默认值
DEFAULT_VALUE = (1, 2, 3) # 默认值，可自行改变
def func(*args):
	if args == (): # 如果为空（用户没有传递参数）:
		args = DEFAULT_VALUE # 设为默认值
	print(args)

In this way, there are default values:

>>> func() # 无参数调用
(1, 2, 3)

to sum up

• *[参数名]Indicates that valueformal parameters should be used , and the number of parameters is not limited. After being passed in, they will be packaged tuplefor use in the function body.
• Special parameter transfer method:*[tuple object]
• This method cannot set the default value, and can only use "Manually set the default value".

---

2. **[Parameter name]

transfer

Legal call

Ordinary call

**参数名Generally written **kwargs, such as:

def func(**kwargs): # kwargs = keyword arguments
	print(kwargs)

Then call func, but this is the opposite of the previous one, it must be a name=valueparameter pass (this is why it is called kwargs (keyword arguments)):

>>> func(a=1, b=2, c=3, d=4)
{
    
    'a': 1, 'b': 2, 'c': 3, 'd': 4}
>>> func(_tuple_obj=tuple(), _set_obj=set(), _dict_obj=dict())
{
    
    '_tuple_obj': (), '_set_obj': set(), '_dict_obj': {
    
    }}
>>> func()
{
    
    }

Such a function can pass any number keyword argument(including 0), here is the opposite of the previous one, the parameters will be packed into one dict, such as {'a': 1, 'b': 2, 'c': 3, 'd': 4} {'_tuple_obj': (), '_set_obj': set(), '_dict_obj': {}} {}.

Special call

What if you already have an dictobject and want to pass kwargsit in?
First define an object like this:

>>> dict_object = {
    
    'a': 666, 'b': 888}
>>> print(dict_object)
{
    
    'a': 666, 'b': 888}

Then, similar to the last time:

>>> func(dict_object) # 因为不能传positional argument, 这下还报错了（马上会讲到）：
Traceback (most recent call last):
  File "<**kwargs test program>", line 1, in <module>
    func(dict_object)
TypeError: func() takes 0 positional arguments but 1 was given
>>> func(**dict_object) # 正确方法
{
    
    'a': 666, 'b': 888}

Illegal call

What if it passes positional argument? Please see:

>>> func(1, 2)

Get TypeError:

Traceback (most recent call last):
  File "<**kwargs test program>", line 1, in <module>
    func(1, 2)
TypeError: func() takes 0 positional arguments but 2 were given

Therefore, the key=valueformal parameter transfer (in English keyword argument) should be used valuehere instead of the parameter transfer method.

Default parameters

Similar to *argsthe method, you should use the method of manually setting the default value:

# 手动设置**kwargs的参数默认值
DEFAULT_VALUE = {
    
    'a': 1, 'b': 2} # 默认值，可自行改变
def func(**kwargs):
	if kwargs == {
    
    }: # 如果为空（用户没有传递参数）:
		kwargs = DEFAULT_VALUE # 设为默认值
	print(kwargs)

In this way, there are default values:

>>> func() # 无参数调用
{
    
    'a': 1, 'b': 2}

to sum up

• **[参数名]Indicates that the key=valueparameters should be passed in form, the number of parameters is not limited, and they will be packaged after being passed in dict.
• Special parameter transfer method:**[dict object]
• This method cannot set the default value, and can only use "Manually set the default value".