appending an element to second copied list does not affect original list

Asked Mar 19 '21 at 14:26

Active Mar 19 '21 at 14:33

Viewed 15 times

lis1 = [1, 2, 3, 4]
lis2 = lis1.copy()
print("The new list created is : " + str(lis2))
lis2.append(5)
print("The new list after adding new element : " + str(lis2))
print("The old list after adding new element to new list  : " + str(lis1))

Output:
The new list created is : [1, 2, 3, 4]
The new list after adding new element : [1, 2, 3, 4, 5]
The old list after adding new element to new list  : [1, 2, 3, 4]

here 5 is appended to list2, but it's not making any changes to list1 Whereas the reverse condition of appending 5 to list1 changes list2 why? doing shallow copy has to change both if the change is done for one, right?

asked Mar 19 '21 at 14:26

muni poojitha

`lis2 = lis1.copy()` makes the changes only to lis2. `lis2 = lis1` makes the changes to both lists. So `...copy()` creates a new object, not referencing to the original object. – Maurice Meyer Mar 19 '21 at 14:29
After you do `lis2 = lis1.copy()` the variables `lis1` and `lis2` are independent, so *"Whereas the reverse condition of appending 5 to list1 changes list2 "* is not true. – Alex Sveshnikov Mar 19 '21 at 14:30
1

That's refreshing... we usually get the opposite question several times a week... – DeepSpace Mar 19 '21 at 14:32
@MauriceMeyer then why changes made to lis1 does affect lis2 if they are two different objects? – muni poojitha Mar 19 '21 at 14:35
@munipoojitha They are not... – DeepSpace Mar 19 '21 at 14:37

appending an element to second copied list does not affect original list

0 Answers0