Python 删除嵌套列表中第二项的引号

我需要删除嵌套列表中第二项的引号。例如，更改：

a = [['first', '41'], ['second', '0'], ['third', '12']]

致：

我试过了

[map(int, [n[1]]) for n in a]
[[41], [0], [12], [0], [45], [17], [3], [10], [1], [19], [98], [0]]

但它删除了第一个元素。感谢您的帮助

[[item[0], int(item[1])] for item in a]

输出：

[['first', 41], ['second', 0], ['third', 12]]

您可以这样做：

a = [['first', '41'], ['second', '0'], ['third', '12']]
a = [[i[0], int(i[1])]for i in a]

>>> print a
[['first', 41], ['second', 0], ['third', 12]]

a = [['first', '41'], ['second', '0'], ['third', '12']]
a = [[i[0], int(i[1])]for i in a]

>>> print a
[['first', 41], ['second', 0], ['third', 12]]