tl;dr: use the copy() method.

Say we have two Python lists – list_a and list_b.

If we try to make a copy of list_a and assign it to list_b using the assignment operator =, what really happens is that both list_a and list_b point to the same memory address.

That means that any list-manipulating actions that are done on either list_a or list_b will affect the same list in memory. We don’t have actually have two separate lists we can act upon.

In the example below, although we append the integer 4 to list_a, we can see that printing out list_b shows the newly added element. That’s because both list variables point to the same memory address:

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list_a = [1, 2, 3]
list_b = []

list_b = list_a

print(list_b) # [1, 2, 3]

list_a.append(4)

print(list_b) # [1, 2, 3, 4]

Output of the program above:

[1, 2, 3]
[1, 2, 3, 4]

To make an actual copy, use the copy() method. Then, when list_a is modified, it is independent of list_b, because list_b is stored in a separate memory address.

Now if we append the same integer 4 to list_a, list_b will be completely unaffected.

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list_a = [1, 2, 3]
list_b = []

list_b = list_a.copy() # using copy()

print(list_b) # [1, 2, 3]

list_a.append(4)

print(list_a) # [1, 2, 3, 4]
print(list_b) # [1, 2, 3]

Output of the program above:

[1, 2, 3]
[1, 2, 3, 4]
[1, 2, 3]

Here’s more proof. We can print out the memory address of each variable to see when they’re the same and when they differ. We can do this using the id() function.

Here are the same lists from above but this time with their unique identifiers printed out. In this case, the IDs match because both list_a and list_b point to the same memory address.

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list_a = [1, 2, 3]
list_b = []

list_b = list_a

print(f'list_a address: {id(list_a)}')
print(f'list_b address: {id(list_b)}')

The program above outputs:

list_a address: 140226819497536
list_b address: 140226819497536

The memory addresses are the same.

Now let’s try the same thing but using the copy() method instead of just an assignment operation with =:

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list_a = [1, 2, 3]
list_b = []

list_b = list_a.copy()

print(f'list_a address: {id(list_a)}')
print(f'list_b address: {id(list_b)}')

The program above outputs:

list_a address: 140264515620480
list_b address: 140264514892160

We can see the memory addresses are different (most obvious due to the ending digits).

Although I’ve been in the field for some time, I still have my smooth brain moments. This is a reminder to myself (and whoever reads this) to remember the basics!

References