列表在嵌套列表中合并

Question

我有以下列表:

[['a', [0, 1]], ['b', [2, 3]], ['c', [4, 5]]]

我想将其更改为:

[['a', 0, 1], ['b', 2, 3], ['c', 4, 5]]`

如何使用列表理解在Python 3中完成？

Answer 1

从Python 3.5+开始,您可以在列表声明中使用解包.在列表理解中使用它:

lst = [['a', [0, 1]], ['b', [2, 3]], ['c', [4, 5]]]

new_lst = [[x, *more] for x, more in lst]

print(new_lst) # [['a', 0, 1], ['b', 2, 3], ['c', 4, 5]]