python - 如何迭代列表中的特定元素？

Question

假设我有一个很大的元组列表，“a”给出如下:

a = [(0,1,2,3,4,5,6,7,8,9), (10, 11, 12, 13, 14, 15, 16, 17,18, 19), (20,21,22,23,24,25,26,27,28).....]

对于这样的列表，如何通过迭代制作另一个列表并组合相应的元素得到以下结果:

a = [(0, 10, 20...), (1, 11, 21)...]

另外，假设我有另一个数组列表，

b = [A、B、C、D、E、F、G、H、I]

那么，我如何组合 a 和 b 来创建另一个列表 c，这样:

c = [A(0,10,20), B(1,11,21) ....]

谢谢!

Answer 1

第一个a列表看起来像是zipped 。您需要解压它。
就像这样 -

>>> a = [(0,1,2,3,4,5,6,7,8,9), (10, 11, 12, 13, 14, 15, 16, 17,18,19), (20,21,22,23,24,25,26,27,28)]
>>> zip(*a)
[(0, 10, 20), (1, 11, 21), (2, 12, 22), (3, 13, 23), (4, 14, 24), (5, 15, 25), (6, 16, 26), (7, 17, 27), (8, 18, 28)]

对于第二个问题，它似乎与第一个问题相反(假设b具有'ABC'作为字符)。所以压缩它们 -

>>> a = zip(*a)
>>> b = "ABCDEFGHI"
>>> zip(b,a)
[('A', (0, 10, 20)), ('B', (1, 11, 21)), ('C', (2, 12, 22)), ('D', (3, 13, 23)), ('E', (4, 14, 24)), ('F', (5, 15, 25)), ('G', (6, 16, 26)), ('H', (7, 17, 27)), ('I', (8, 18, 28))]

编辑
正如 gnibbler 在评论中所建议的，此方法涉及参数的打包 - (假设 b 作为一个复杂的列表)

>>> b = [[65, 66], [67, 68], [69, 70], [71, 72], [73, 74], [75, 76], [77, 78], [79, 80], [81, 82]]
>>> zip(b, *a)
[([65, 66], 0, 10, 20), ([67, 68], 1, 11, 21), ([69, 70], 2, 12, 22), ([71, 72], 3, 13, 23), ([73, 74], 4, 14, 24), ([75, 76], 5, 15, 25), ([77, 78], 6, 16, 26), ([79, 80], 7, 17, 27), ([81, 82], 8, 18, 28)]

为了更清楚地了解差异，这是我之前的答案的示例 -

>>> zip(b, zip(*a))
[([65, 66], (0, 10, 20)), ([67, 68], (1, 11, 21)), ([69, 70], (2, 12, 22)), ([71, 72], (3, 13, 23)), ([73, 74], (4, 14, 24)), ([75, 76], (5, 15, 25)), ([77, 78], (6, 16, 26)), ([79, 80], (7, 17, 27)), ([81, 82], (8, 18, 28))]