Python入门之Python函数

'earth.mars.venus'
>>> concat("earth", "mars", "venus", ".")

此外，如果我们有一个列表

["earth", "mars", "venus"]

，或者元组

("earth", "mars", "venus")

，想传入 concat()函数怎么办呢？如果我们直接把列表或者元组传进去都会报错，正确的做法是解包参数列表，即传入

*List

和

*tuple

的形式，

*

可以理解为去掉

[]

或者

()

，这非常常用且实用。示例如下：

>>> list_a = ["earth", "mars", "venus"]>>> concat(list_a)
Traceback (most recent call last):
  File "", line 1, in
  File "", line 2, in concat
TypeError: sequence item 0: expected str instance, list found
>>> concat(*list_a)
'earth/mars/venus'
>>> t = ("earth", "mars", "venus")
>>> concat(t)
Traceback (most recent call last):
  File "", line 1, in
  File "", line 2, in concat
TypeError: sequence item 0: expected str instance, tuple found
>>> concat(*t)
'earth/mars/venus'

**kwargs

可变关键字参数示例：

def kwargs_fun(**kwargs):
    for key in kwargs:
        print(f'{key}：{kwargs[key]}')
kwargs_fun(a=1, b=2, c=3)
输出结果：
a：1
b：2
c：3

*args

和

**kwargs

组合使用示例：

def cheeseshop(kind, *arguments, **keywords):
    print("-- Do you have any", kind, "?")
    print("-- I'm sorry, we're all out of", kind)
    for arg in arguments:
        print(arg)
    print("-" * 40)
    for kw in keywords:
        print(kw, ":", keywords[kw])
cheeseshop("Limburger", "It's very runny, sir.",
           "It's really very, VERY runny, sir.",
           shopkeeper="Michael Palin",
           client="John Cleese",
           sketch="Cheese Shop Sketch")
输出结果：
-- Do you have any Limburger ?
-- I'm sorry, we're all out of Limburger
It's very runny, sir.
It's really very, VERY runny, sir.
----------------------------------------
shopkeeper : Michael Palin
client : John Cleese
sketch : Cheese Shop Sketch

此外，字典中也可以用解包，符号是

**dict

，相当于去掉

{}

，测试如下：

>>> def parrot(voltage, state='a stiff', action='voom'):
 10/16   首页 上一页 8 9 10 11 12 13 下一页 尾页