Python Question / Quiz; What is the output of the following Python code, and why? Comment your answers below!

@PythonPr y = x.copy() -->Creates a copy of list x x.append(4) -->Adds 4 to list x print(y) -->Prints y, which remains unchanged. x becomes [1,2,3,4] y remains [1,2,3] --> Ans. a

@PythonPr The correct answer is (a) [1, 2, 3]. The code y = x.copy() creates a shallow copy of the list x. This means that y is a new list object with the same elements as x, but it is a distinct object in memory. When x.append(4) is called, it modifies the original list x by

@PythonPr The .copy() method creates a separate reality. y gets a snapshot of x's state, then their paths diverge. When x is updated with 4, y is oblivious, frozen in its original moment. It’s about a point in time, not a permanent link. That's why y is [1, 2, 3].

@PythonPr A. [1, 2, 3]. The append method is only binding on x. So y remains.

@PythonPr Comprobado: a

@PythonPr a, no harm to y

@PythonPr Is there any test that would make you want to open an IDE to run it?🤔 (Answer "A")😒

@PythonPr >>> x = [1,2,3] >>> y = x.copy() >>> x.append(4) >>> print(y) [1, 2, 3] >>> print(x) [1, 2, 3, 4] >>>

@PythonPr a for sure

@PythonPr Alternative A: [1,2,3] The varables points to different list at memory address. The y list is a copy, but new, x list. This means that alterations isn't shared between the lists.

@PythonPr A.

@PythonPr A)

@PythonPr A

@PythonPr B

@PythonPr A

@PythonPr B

@PythonPr a

@PythonPr a

@PythonPr B

@PythonPr a,[1,2,3]

@PythonPr a

@PythonPr B

@PythonPr 123？

@PythonPr A

@PythonPr A

@PythonPr C

@PythonPr the answer is B, because of the ".append" method, it adds the new "item" to the tail/end of the list

@PythonPr B

@PythonPr [1, 2, 3]

@PythonPr c

@PythonPr A

@PythonPr A

@PythonPr C

@PythonPr A [1,2,3] karena y adalah fotokopi dari x. Ketika x ditambahkan (4), cuma ada di kertas x, tp tidak difotokopian y. So, the result still same: y=x.copy(), whichis [1,2,3].

@PythonPr A

@PythonPr A ✨ y is a copy of x, not a reference to the same array