問題：讀取movies = ['a','b',['c','d',['f','g','h']]]里的每個字符串

方法：利用isinstance、for和if...else讀取列表里的字符串。

具體步驟：

z =0
k = 0
q = 0
movies = ['a','b',['c','d',['f','g','h']]]
for each_item in movies:
    if isinstance(each_item,list)==False:
        print(each_item)
        z = z +1
    else:
        for p in each_item:
            if isinstance(p,list)==False:
                print(movies[z][k])
                k = k+1
            else:
                for s in p:
                    print(movies[z][k][q])
                    q=q+1

結(jié)果：
可是解決問題

存在問題：
如果嵌套級別太多，就要重復使用for和if...else，不斷增加變量。

進一步優(yōu)化：

movies = ['a','b',['c','d',['f','g','h']]]
for each_item in movies:
    if isinstance(each_item,list):
        for nested_item in each_item:
            if (nested_item, list) == False:
                print(nested_item)
            else:
                for f in nested_item:
                    print(f)
    else:
        print(each_item)

或者

movies = ['a','b',['c','d',['f','g','h']]]
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:
                    print(deeper_item)
            else:
                 print(nested_item)
    else:
        print(each_item)

總結(jié)：代碼更少，清晰，但還存在一個問題：如果有更多的嵌套，該如何解決。
比如說movies = ['a','b',['c','d',['f','g','h',['o']]]]

使用函數(shù)，繼續(xù)優(yōu)化形成最終大法：

movies = ['a','b',['c','d',['f','g','h',['o']]]]
def item (name):
    for each_item in name:
        if isinstance(each_item,list):
            item(each_item)
        else:
            print(each_item)
print(item(movies))

這樣，再也不怕嵌套的級別了。

問題解決方法參考了《Head First Python(中文版)》