-- coding: utf-8 --

展开正则表达式用循环分析:

L1 = ['Hello', 'World', 18, 'Apple', None] L2 = []

for x in L1 : if isinstance(x , str): L2.append(x) else: pass print(L2) print([s.lower() for s in L2])

print([s.lower() for s in L1 if isinstance(s,str)])

关于s.lower() for s in L2 循环表现为: s.lower() for s in L2: L3.append(s) 假定L3是一个空List.

#2 Created at ... [Delete] [Delete and Lock User]

补充一下,在原题中已经有了print(L2) 所以在答题上填: L2 = []

for x in L1 : if isinstance(x , str): L2.append(x) else: pass

L2 = [s.lower() for s in L2] 即可

#3 Created at ... [Delete] [Delete and Lock User]

L2 = [s.lower() for s in L1 if isinstance(s, str)] 上述是正确答案