10.4 Modifying Lists

All of these mutate the list in place — the list’s identity (id()) stays the same throughout; see 10.11 Copying and Mutability.

Update

Assign directly to an index to replace that single element.

>>> l = [1, 2, 3]
>>> l[0] = 99
>>> l
[99, 2, 3]

Append

Adds one item to the end — the most common way to grow a list.

>>> l.append(4)
>>> l
[99, 2, 3, 4]

Extend

Adds each item from another iterable individually — different from append(), which would add the whole iterable as one nested element.

>>> l.extend([5, 6])
>>> l
[99, 2, 3, 4, 5, 6]

Insert

Adds an item at a specific position, shifting everything after it one slot to the right — see 10.13 Performance for why this is more expensive than append().

>>> l.insert(1, 100)
>>> l
[99, 100, 2, 3, 4, 5, 6]

Replace (Slice Assignment)

Assigning to a slice replaces a whole range of elements at once, and the replacement can even be a different length than the original range.

>>> l[1:3] = [7, 8, 9]
>>> l
[99, 7, 8, 9, 3, 4, 5, 6]

Quick Interview Answer

“Every modification method here mutates the list in place rather than returning a new one — the list’s id() never changes. l[i] = x replaces a single element. append(x) adds exactly one item to the end; extend(iterable) unpacks and adds each item individually — a very common point of confusion, since append([1, 2]) nests a whole list as one element instead of adding 1 and 2 separately. insert(i, x) shifts everything after position i to make room. Slice assignment is the most powerful of the group — it can replace, grow, or shrink a range in one statement, since the replacement doesn’t need to match the original range’s length.”

Common Mistakes

• Calling l.append([1, 2]) when l.extend([1, 2]) was intended — append nests the whole list as a single element, producing [..., [1, 2]] instead of adding 1 and 2 individually.
• Using l.insert(0, x) repeatedly in a loop instead of building the list in order and calling append() — each insert(0, ...) is O(n), turning the loop into O(n²).
• Forgetting slice assignment’s replacement length doesn’t need to match the slice’s length — l[1:3] = [7, 8, 9] replaces two elements with three, growing the list.