1 >>> movies=["The Holy Grail", 1975, "The Life of Brian", 1979, "The Meaning of Life", 1983] 2 >>> for eachMovie in movies: 3 print(eachMovie)
按下两个回车后输出结果如下：
1 The Holy Grail 2 1975 3 The Life of Brian 4 1979 5 The Meaning of Life 6 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
1 >>> def print_lol(movies): 2 for each_item in movies: 3 if isinstance(each_item,list): 4 print_lol(each_item) 5 else: 6 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：
1 The Holy Grail 2 1975 3 Terry Jones & Terry Gilliam 4 91 5 Graham Chapman 6 Michael Palin 7 John Cleese 8 Terry Gilliam 9 Eric Idle 10 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：
1 a 2 b 3 1 4 3 5 hello 6 22 7 33 8 ok 9 h 10 zhuangshi
再看这个函数：
1 >>> def print_lol(movies): 2 for each_item in movies: 3 if isinstance(each_item,list): 4 print_lol(each_item) 5 else: 6 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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