Updating a sliced list

Question

I thought I understood Python slicing operations, but when I tried to update a sliced list, I got confused:

>>> foo = [1, 2, 3, 4]
>>> foo[:1]          


        

Answer 1

This is because python does not have l-values that could be assigned. Instead, some expressions have an assignment form, which is different.

A foo[something] is a syntactic sugar for:

foo.__getitem__(something)

but a foo[something] = bar is a syntactic sugar for rather different:

foo.__setitem__(something, bar)

Where a slice is just a special case of something, so that foo[x:y] expands to

foo.__getitem__(slice(x, y, None))

and foo[x:y] = bar expands to

foo.__setitem__(slice(x, y, None), bar)

Now a __getitem__ with slice returns a new list that is a copy of the specified range, so modifying it does not affect the original array. And assigning works by the virtue of __setitem__ being a different method, that can simply do something else.

However the special assignment treatment applies only to the outermost operation. The constituents are normal expressions. So when you write

foo[:][1] = 'two'

it gets expanded to

foo.__getitem__(slice(None, None, None)).__setitem__(1, 'two')

the foo.__getitem__(slice(None, None, None)) part creates a copy and that copy is modified by the __setitem__. But not the original array.