You can pass a list as an argument of a Python function the same way as any other type and it will be the same type inside the function.
Take a look at the following example:
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my_list = [1, 2, 3] def my_func(list_as_argument): for elem in list_as_argument: print(elem) my_func(my_list) print(my_list) |
There is a list of three numbers. You can pass this list as an argument and display all its elements one by one:
1 2 3 [1, 2, 3]
If you print a list after the function execution, it will print the same numbers.
Because lists are mutable, when you pass a list as an argument, you create a reference to this list and not a copy.
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my_list = [1, 2, 3] def my_func(list_as_argument): list_as_argument[0] = 8 print(list_as_argument) print(my_list) my_func(my_list) print(my_list) |
If you print the list it will display numbers: 1, 2, 3.
After you modify this list, the modification will apply to the list outside the function because it was a reference and not a copy.
[1, 2, 3] [8, 2, 3] [8, 2, 3]
Creating a copy of a list
If you want to modify the list as a copy of the one inside parameter, you can use the copy function.
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my_list = [1, 2, 3] def my_func(list_as_argument): new_list = list_as_argument.copy() new_list[0] = 8 print(new_list) print(my_list) my_func(my_list) print(my_list) |
Now, if you run the code, the modification inside the function doesn’t apply to the my_list list, because the function operates on a copy and not on the original object.
[1, 2, 3] [8, 2, 3] [1, 2, 3]
