If we pass list as parameter to a function and change the parameter, the original list is also changed. This is because list is a mutable type, when we pass list to a function, we are passing the same list.

def my_fun(my_list):
    my_list.append(1)

    return my_list


x = [1, 2, 3]
y = my_fun(x)

print(f"x: {x}, y: {y}")
# x, y are both [1, 2, 3, 1]

How can we pass the “value” of this list instead of its “reference”? We can use several ways:

• slicing: my_fun(x[:])
• list(): my_fun(list(x))
• list.copy(): my_fun(x.copy())
• copy.copy(): my_fun(copy.copy(x))
• copy.deepcopy(): my_fun(copy.deepcopy(x))

The first four ways only create a shallow copy of the original list. They only work for simple list consisting of immutable types, for example, a list of int. If the list element is a mutable type themselves, they will not work.

Only the copy.deepcopy() method can truly create a new list.

references#

• https://docs.python.org/3/faq/programming.html#why-did-changing-list-y-also-change-list-x
• clone a list so that it doesn’t change after assignment: https://stackoverflow.com/q/2612802/6064933
• pass list as argument: https://stackoverflow.com/q/2322068/6064933
• pass list to function by value: https://stackoverflow.com/q/15377050/6064933