Python Tutorial / Create Lists and Add, Change and Remove Items
A list stores ordered items and allows changes. Distinguish operations that mutate the original list from functions that return a new list, especially when sorting or copying values.
Choose the operation by its purpose
Create a list with square brackets and access its first item at index zero. Lists can contain duplicates and different types, but values used in a total should have a consistent meaning and unit.
| Operation | Code | Important behavior |
|---|---|---|
| Append one item | items.append(value) | A list argument becomes one nested item |
| Append several items | items.extend(values) | Adds each item from an iterable |
| Remove by position | items.pop(index) | Returns the removed item |
| Remove by value | items.remove(value) | Removes the first match; fails if absent |
For the relevant APIs and platform requirements, see Python: datastructures.
List sharing and transformation build on the rules demonstrated in Build List, Dictionary and Set Comprehensions.
Run list changes and sorting
Save the complete code below as lists.py in a separate practice folder, using UTF-8. Run python lists.py in a terminal with your project virtual environment selected. Outside a virtual environment on macOS or Ubuntu, use python3 when that is your interpreter command.
amounts = [3500, 8000, 2500] amounts.append(1000) amounts[0] = 4000 removed = amounts.pop() print(amounts, removed) print(sorted(amounts)) print(amounts) amounts.sort() print(amounts)
Expected output:
[4000, 8000, 2500] 1000 [2500, 4000, 8000] [4000, 8000, 2500] [2500, 4000, 8000]
Separate mutation, sorting and copying
sorted(amounts) returns a new list and preserves the original order. amounts.sort() changes the existing list and returns None. Writing amounts = amounts.sort() therefore loses the list reference by assigning None.
A slice such as amounts[:] creates a shallow copy of the outer list. Nested mutable objects remain shared. A simple assignment other = amounts does not even copy the outer list; it creates another name for the same object. The object and copy lesson explores these effects.
list.sort() sorts the original list and returns None; sorted(iterable) returns a new list. Consequently, items = items.sort() assigns None rather than a sorted list. Neither operation makes a deep copy of nested objects.
| Operation | Result | Observation |
|---|---|---|
| items.sort() | None | Changes original |
| sorted(items) | New list | Original order retained |
Handle empty data and changes during iteration
Reading index zero or calling pop() on an empty list raises IndexError. Decide whether an empty input is valid or violates a requirement. remove() raises ValueError for an absent value.
Removing items from the list you are currently iterating can skip items as positions shift. Iterate over a copy or construct a new filtered list instead. Add zero to the original amounts, inspect both order and total, then restore the data. A change that preserves a sum can still change the structure.
