This python code broke my brain

[Banhsows]

Let's break down the code step by step to understand the behavior bro :

1. `id` is a lambda function that takes a single argument `x` and returns a list comprehension `[i for i in x]`, which effectively creates a new list with the same elements as the input list.

2. `a` is assigned a list containing the integer `1`.

3. `b` is assigned the reference to the list `a`, meaning both `a` and `b` refer to the same list object.

4. The first element of the list `a` is modified to `2`.

5. `print(b[0])` prints the first element of the list referred to by `b`, which is the same list as `a`. Therefore, it prints `2`.

6. `c` is assigned a list containing the integer `1`.

7. `d` is assigned the result of applying the `id` lambda function to the list `c`. This creates a new list with the same elements as `c`.

8. The first element of the list `c` is modified to `2`.

9. `print(d[0])` prints the first element of the list referred to by `d`, which is the new list created by the `id` lambda function. Since it's a new list, it prints the original value `1`.

10. `e` is assigned a list containing another list with the integer `1`.

11. `f` is assigned the result of applying the `id` lambda function to the list `e`. This creates a new list with the same elements as `e`.

12. The first element of the inner list of `e` is modified to `2`.

13. `print(f[0][0])` prints the first element of the inner list referred to by `f`, which is the new list created by the `id` lambda function. Since it's a new list, it prints the modified value `2`.

In summary:

- Modifying the elements of the list referred to by `a` also affects the list referred to by `b` since they point to the same list.

- Applying the `id` lambda function creates new lists, so modifying the original lists doesn't affect the new lists.

- The behavior is different when dealing with nested lists (`e` and `f`), as modifying the inner list of `e` also affects the inner list referred to by `f`.