问题

my_tuples = [(1, 2, 3), ('a', 'b', 'c', 'd', 'e'), (True, False), 'qwerty']
这个元组，如何将所有值附加到列表和字典中。例如，我希望得到类似于list1 = [1,2,3,a,b,c] 的输出，以此类推，对于字典，我希望得到类似{1:1,2:2,3:3,4:a} 的输出。

对于列表，我尝试使用列表的长度，但是列表元组中的元素长度都不同。因此，我会得到索引超出范围的错误。
我希望将元组列表的所有元素作为单个列表元素。

答案1

得分: -1

使用嵌套的推导式，首先迭代`my_tuples`中的元组，然后迭代每个元组中的项，将它们放入一个单一的列表或字典中：

```python
>>> my_tuples = [(1, 2, 3), ('a', 'b', 'c', 'd', 'e'), (True, False), 'qwerty']
>>> [i for t in my_tuples for i in t]
[1, 2, 3, 'a', 'b', 'c', 'd', 'e', True, False, 'q', 'w', 'e', 'r', 't', 'y']
>>> {i: v for i, v in enumerate((i for t in my_tuples for i in t), 1)}
{1: 1, 2: 2, 3: 3, 4: 'a', 5: 'b', 6: 'c', 7: 'd', 8: 'e', 9: True, 10: False, 11: 'q', 12: 'w', 13: 'e', 14: 'r', 15: 't', 16: 'y'}

由于'qwerty'不是一个元组，它被解包为其各个字符（因为字符串是字符的可迭代对象，就像(1, 2, 3)是整数的可迭代对象一样）。如果这不是你想要的，你可以将它放入你的元组列表中，如(qwerty,)，使其成为一个单元素元组，或者你可以在推导式中使用类似(t if isinstance(t, tuple) else (t,))的表达式来完成。

<details>
<summary>英文:</summary>

Use a nested comprehension to iterate first over the tuples in `my_tuples` and then the items in each tuple, putting them into a single list or dictionary:

>>> my_tuples = [(1, 2, 3), ('a', 'b', 'c', 'd', 'e'), (True, False), 'qwerty']
>>> [i for t in my_tuples for i in t]
[1, 2, 3, 'a', 'b', 'c', 'd', 'e', True, False, 'q', 'w', 'e', 'r', 't', 'y']
>>> {i: v for i, v in enumerate((i for t in my_tuples for i in t), 1)}
{1: 1, 2: 2, 3: 3, 4: 'a', 5: 'b', 6: 'c', 7: 'd', 8: 'e', 9: True, 10: False, 11: 'q', 12: 'w', 13: 'e', 14: 'r', 15: 't', 16: 'y'}

Since `'qwerty'` isn't a tuple, it's unpacked into its individual characters (since a string is an iterable of characters, the same way that `(1, 2, 3)` is an iterable of ints).  If that wasn't what you were looking for, you could put it in your list of tuples as `(qwerty,)` to make it a one-element tuple, or you could do it inside the comprehension with an expression like `(t if isinstance(t, tuple) else (t,))`.


</details>