python - Changing list elements in shallow copy

Question

I have one question about list shallow copy.

In both examples, I modified one element of the list, but in example 1, list b changed, while in example 2, list d is not changed. I am confused since in both examples, I modified an element of the list.

What's the difference?

Example 1:

a=[1,2,[3,5],4]
b=list(a)
a[1]=0
print(a)   # [1, 0, [3, 5], 4]
print(b)   # [1, 2, [3, 5], 4]

Example 2:

c=[1,2,[3,5],4]
d=list(c)
c[2][0]=0
print(c)   # [1, 2, [0, 5], 4]
print(d)   # [1, 2, [0, 5], 4]

Answer 1

A shallow copy means that you get a new list but the elements are the same. So both lists have the same first element, second element, etc.

If you add, remove, or replace a value from the shallow copied list that change is not reflected in the original (and vise-versa) because the shallow copy created a new list. However if you change an element in either that change is visible in both because both lists reference the same item. So the inner list is actually shared between both the new list and the old list and if you change it, that change is visible in both.

Note that you actually didn't change an element in either example, you replace an element of the list in the first example and in the second example, you replace an element of an element of your list.

I'm currently using graphviz a lot so let me add some images to illustrate this:

The shallow copy means you get a new list but the objects stored in the list are the same:

If you replace an element in any of these the corresponding element will just reference a new item (your first example). See how one list references the two and the other the zero:

While a change to an referenced item will change that item and every object that references that item will see the change: