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)))

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)

这可以通过列表推导来完成

>>> [x.lower() for x in ["A", "B", "C"]]
['a', 'b', 'c']
>>> [x.upper() for x in ["a", "b", "c"]]
['A', 'B', 'C']

或者使用映射函数

>>> list(map(lambda x: x.lower(), ["A", "B", "C"]))
['a', 'b', 'c']
>>> list(map(lambda x: x.upper(), ["a", "b", "c"]))
['A', 'B', 'C']

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

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']

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