What does List[:] = List do in this radix sort code?

Question

Here is a link to a radix sort code i found. It works (the top answer's code) but my question is what does List[:] = [] do in this context (I know its meant for list splitting). I mean you suppose to reconstruct the list from the buckets after every iteration but when I change that line to List = [] the code stops working. Why is that so?

score 1 · Accepted Answer · answered Aug 12 '17 at 07:23

Consider the following case, which is a small version of what you are trying to understand:

def foo(x):
    x = []
    x.append(1)
    print(x)

my_list = []
foo(my_list)
print(my_list)

You expect the function foo to clear the list it received as an argument and, afterward, modify it by appending the number 1, right? Actually, that is not what happens...

>>> foo(my_list)
[1]
>>> print(my_list)
[]

Why so?

Because when you wrote x = [], instead of clearing the original list, you actually changed the local variable x to refer to a new list. On the other hand, my_list is still referring to the original list!

Now, let's try the same but using x[:] instead:

def foo(x):
    x[:] = []
    x.append(1)
    print(x)

my_list = []
foo(my_list)
print(my_list)

Now, everything should work as you expected:

>>> foo(my_list)
[1]
>>> print(my_list)
[1]

The x[:] trick is basically telling the interpreter to replace the contents of the list referred by x by a new empty list. Now, it is not x that will change to which list it is pointing to. In fact, it will still refer to the same list as my_list and, as a side effect, this will alter the original list.

In order to truly grasp what is going on behind the scenes, take a look at how variables are passed by assignment in Python, specially if you aren't familiar with the concept of pointers.

What does List[:] = List do in this radix sort code?

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