Shallow Copy VS. Deep Copy in Python

Question

On a website this statement is mentioned: "In case of shallow copy, a reference of object is copied in other object. It means that any changes made to a copy of object do reflect in the original object."
But with my example this doesn't work
old_list = [1,3,5]
new_list = old_list.copy()
new_list.append(8)
print(old_list)
print(new_list)

OUTPUT:
[1, 3, 5]
[1, 3, 5, 8]

Answer 1

~ swim ~
That seems to work but if I change the whole inner list then this doesn't work.
old_list = [1,3,[5,6]]
new_list = old_list.copy()
new_list[2]= [2,0]
print(old_list)
print(new_list)

OUTPUT:
[1, 3, [5, 6]]
[1, 3, [2, 0]]

Why?

Answer 2

I agree with what ~ swim ~ said. In your second example, the object at index 2 in new_list is being replaced. The object itself is not modified, only the reference is lost. So new_list[2] no longer points to the same object. The address also changes, as can be checked using id().

old_list = [1,3,[5,6]]
new_list = old_list.copy()
new_list[2].append(7)
print(id(old_list[2]), id(new_list[2]))
new_list[2]= [2,0]

print(id(old_list[2]), id(new_list[2]))

Answer 3

Kishalaya Saha please see this.