Python: Pass By Reference And Slice Assignment

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) ... &

Solution 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:

deff(a):
    print(a)
    b = a[:2]
    print(b)

Instead, you should be changing ain place such as:

deff(a):
    print(a)
    a[:] = a[:2]
    print(a)

Solution 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:]
# ora[:] = 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.

Solution 3:

Indeed the objects are passed by reference but a = a[:2] basically creates a new local variable that points to slice of the list.

To modify the list object in place you can assign it to its slice(slice assignment).

Consider a and b here equivalent to your global b and local a, here assigning a to new object doesn't affect b:

>>>a = b = [1, 2, 3]    >>>a = a[:2]  # The identifier `a` now points to a new object, nothing changes for `b`.>>>a, b
([1, 2], [1, 2, 3])
>>>id(a), id(b)
(4370921480, 4369473992)  # `a` now points to a different object

Slice assignment work as expected:

>>>a = b = [1, 2, 3]    >>>a[:] = a[:2]  # Updates the object in-place, hence affects all references.>>>a, b
([1, 2], [1, 2])
>>>id(a), id(b)
(4370940488, 4370940488)  # Both still point to the same object

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Related: What is the difference between slice assignment that slices the whole list and direct assignment?