Python: Pass by reference and slice assignment

Question

In Python, lists are passed by reference to functions, right?

If that is so, what's happening here?

>>> def f(a):
...     print(a)
...     a = a[:2]
...     print(a)
...
>>> b = [1,2,3]
>>> f(b)
[1, 2, 3]
[1, 2]
>>> print(b)
[1, 2, 3]
>>>

Answer 1

In the statement:

a = a[:2]

you are creating a new local (to f()) variable which you call using the same name as the input argument a.

That is, what you are doing is equivalent to:

def f(a):
    print(a)
    b = a[:2]
    print(b)

Instead, you should be changing a in place such as:

def f(a):
    print(a)
    a[:] = a[:2]
    print(a)

Answer 2

When you do:

a = a[:2]

it reassigns a to a new value (The first two items of the list).

All Python arguments are passed by reference. You need to change the object that it is refered to, instead of making a refer to a new object.

a[2:] = []
# or
del a[2:]
# or
a[:] = a[:2]

Where the first and last assign to slices of the list, changing the list in-place (affecting its value), and the middle one also changes the value of the list, by deleting the rest of the elements.