Use my_list.pop(index) to remove an item by its position and keep the removed value, or del my_list[index] to delete it without returning a value. Python list indexes start at 0, so index 0 means the first item.
Remove an item by index with pop() or del
For example, these two forms both remove "banana" from the list, but only pop() gives you the removed item:
items = ["apple", "banana", "cherry"]
removed = items.pop(1)
# items is ["apple", "cherry"]
# removed is "banana"items = ["apple", "banana", "cherry"]
del items[1]
# items is ["apple", "cherry"]The Python tutorial documents both approaches: Python 3.14.8: Data Structures.
Choose the operation that matches your goal
| Operation | What it does | Use it when |
|---|---|---|
my_list.pop(i) |
Removes and returns the item at index i. |
You need the removed item as well as the shorter list. |
del my_list[i] |
Deletes the item at index i without returning it. |
You only need to change the list. del is a Python statement, not a list method. |
my_list.remove(value) |
Searches for the first item equal to value and removes it. |
The target is specified by its value, not its position. It raises ValueError if no matching item exists. |
For example, items.remove("banana") searches by value; it does not mean “remove the item at index 1.”
Free tools Windows power users keep installed
One-click scans. No signup required.
#1 Best Overall
Understand indexes and invalid positions
Indexes begin at zero: index 0 is the first item and index 1 is the second. A negative index counts from the end. Calling items.pop() without an argument removes and returns the last item.
pop raises IndexError if the list is empty or the index is outside the valid range. If an invalid index is an expected condition in your program, handle that case explicitly; otherwise, validating the position or letting the exception reveal a bug may be more appropriate.
Remove items at several indexes
Each deletion changes the positions of items that follow it. If you need to remove several known indexes from the same list, remove them in descending order so deleting a lower position does not shift the higher positions you still intend to delete:
items = ["a", "b", "c", "d"]
indexes = [1, 3]
for index in sorted(indexes, reverse=True):
del items[index]
# items is ["a", "c"]If the items to remove are defined by a condition rather than by fixed positions, building a new list with a filter is often clearer than repeatedly deleting indexes.
What’s actually slowing this PC down?
Pick the symptom - the matching free tool is one click away.
What repeated indexed deletion costs
Deleting an item near the beginning of a list may require shifting later items. The CPython built-in types complexity reference gives indexed pop and item deletion a cost of O(n – k), where n is the current list size and k is the index: CPython built-in types time complexity reference. For a single ordinary deletion, pop or del is the straightforward choice; if your program frequently adds or removes items at both ends, the reference recommends considering collections.deque.
Quick Recap
Product prices and availability are accurate as of the date/time indicated and are subject to change. Any price and availability information displayed on Amazon at the time of purchase will apply.
