问题：读取movies = ['a','b',['c','d',['f','g','h']]]里的每个字符串
方法：利用isinstance、for和if...else读取列表里的字符串。
具体步骤：
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
结果：
可是解决问题
存在问题：
如果嵌套级别太多，就要重复使用for和if...else，不断增加变量。
进一步优化：
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)
总结：代码更少，清晰，但还存在一个问题：如果有更多的嵌套，该如何解决。
比如说movies = ['a','b',['c','d',['f','g','h',['o']]]]
使用函数，继续优化形成最终大法：
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))
这样，再也不怕嵌套的级别了。
问题解决方法参考了《Head First Python(中文版)》
