>>> movies=["The Holy Grail", 1975, "The Life of Brian", 1979, "The Meaning of Life", 1983]
>>> for eachMovie in movies:
print(eachMovie)
按下两个回车后输出结果如下：
The Holy Grail
1975
The Life of Brian
1979
The Meaning of Life
1983
列表中还可以存储其他列表！
movies = ["The Holy Grail", 1975, "Terry Jones & Terry Gilliam", 91,
["Graham Chapman",
["Michael Palin", "John Cleese", "Terry Gilliam", "Eric Idle", "Terry Jones"]
]
]
这样的嵌套列表用上面的for循环怎么处理呢？
>>> for each_item in movies:
print(each_item) The Holy Grail
1975
Terry Jones & Terry Gilliam
91
['Graham Chapman', ['Michael Palin', 'John Cleese', 'Terry Gilliam', 'Eric Idle', 'Terry Jones']]
哦！没有处理干净......
再循环！
>>> for each_item in movies:
if isinstance(each_item,list):
for nested_item in each_item:
if isinstance(nested_item,list):
for deeper_item in nested_item:
if isinstance(deeper_item,list):
for deepest_item in deeper_item:
print(deepest_item)
else:
print(deeper_item)
else:
print(nested_item)
else:
print(each_item) The Holy Grail
1975
Terry Jones & Terry Gilliam
91
Graham Chapman
Michael Palin
John Cleese
Terry Gilliam
Eric Idle
Terry Jones
哇塞!这倒是处理干净了，但是也太绕了吧！不过队形不错！
用函数来处理
def 函数名（参数）: 函数代码组
>>> def print_lol(movies):
for each_item in movies:
if isinstance(each_item,list):
print_lol(each_item)
else:
print(each_item)
6行代码即可完成上面的晕头转向。
引用函数：
>>> print_lol (movies)
结果如下：
The Holy Grail
1975
Terry Jones & Terry Gilliam
91
Graham Chapman
Michael Palin
John Cleese
Terry Gilliam
Eric Idle
Terry Jones
我们以后就可以用这个函数来处理各种嵌套列表了。
定义一个列表：
>>> hello=["a","b",1,3,["hello",22,33,["ok","h"]],"zhuangshi"]
调用函数：
>>> print_lol(hello)
结果如下：
a
b
1
3
hello
22
33
ok
h
zhuangshi
再看这个函数：
>>> def print_lol(movies):
for each_item in movies:
if isinstance(each_item,list):
print_lol(each_item)
else:
print(each_item)
在代码内部引用了自身！——递归！没错是递归。Python3默认递归深度不能超过100，但是这个深度上限可以改。
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