mylist = ['Mixed Case One', 'Mixed Case Two', 'Mixed Three']
print(list(map(lambda x: x.lower(), mylist)))
print(list(map(lambda x: x.upper(), mylist)))

如果你的目的是通过一次转换来匹配另一个字符串，你也可以使用str.casefold()。

这是有用的，当你有非ascii字符和匹配ascii版本(例如:maße vs masse)。虽然str.lower或str.upper在这种情况下失败，但str.casefold()将通过。 这在Python 3中可用，详细讨论了这个想法，答案是https://stackoverflow.com/a/31599276/4848659。

>>>str="Hello World";
>>>print(str.lower());
hello world
>>>print(str.upper());
HELLO WOLRD
>>>print(str.casefold());
hello world

>>> list(map(str.lower,["A","B","C"]))
['a', 'b', 'c']

对于这个例子，理解是最快的

$ python -m timeit -s 's=["one","two","three"]*1000' '[x.upper for x in s]'
1000 loops, best of 3: 809 usec per loop

$ python -m timeit -s 's=["one","two","three"]*1000' 'map(str.upper,s)'
1000 loops, best of 3: 1.12 msec per loop

$ python -m timeit -s 's=["one","two","three"]*1000' 'map(lambda x:x.upper(),s)'
1000 loops, best of 3: 1.77 msec per loop

mylist = ['Mixed Case One', 'Mixed Case Two', 'Mixed Three']
print(list(map(lambda x: x.lower(), mylist)))
print(list(map(lambda x: x.upper(), mylist)))

除了更容易阅读(对许多人来说)，列表推导式也在速度竞赛中获胜:

$ python2.6 -m timeit '[x.lower() for x in ["A","B","C"]]'
1000000 loops, best of 3: 1.03 usec per loop
$ python2.6 -m timeit '[x.upper() for x in ["a","b","c"]]'
1000000 loops, best of 3: 1.04 usec per loop

$ python2.6 -m timeit 'map(str.lower,["A","B","C"])'
1000000 loops, best of 3: 1.44 usec per loop
$ python2.6 -m timeit 'map(str.upper,["a","b","c"])'
1000000 loops, best of 3: 1.44 usec per loop

$ python2.6 -m timeit 'map(lambda x:x.lower(),["A","B","C"])'
1000000 loops, best of 3: 1.87 usec per loop
$ python2.6 -m timeit 'map(lambda x:x.upper(),["a","b","c"])'
1000000 loops, best of 3: 1.87 usec per loop