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

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

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

如果你试图将列表中的所有字符串转换为小写，你可以使用pandas:

import pandas as pd

data = ['Study', 'Insights']

pd_d = list(pd.Series(data).str.lower())

输出:

['study', 'insights']

一个学生问，另一个有同样问题的学生回答:)

fruits=['orange', 'grape', 'kiwi', 'apple', 'mango', 'fig', 'lemon']
newList = []
for fruit in fruits:
    newList.append(fruit.upper())
print(newList)

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)))