IndexError: list index out of range (Fix Explained)

IndexError: list index out of range means your code is trying to read a list position that does not exist.

This usually happens when:

the index is too large
the list is empty
a loop goes one step too far
you expect more items than are really in the list

The good news is that this error is usually easy to fix. In most cases, you need to check the list length with len() or use a safer looping pattern.

Quick fix

items = ["a", "b", "c"]
index = 2

if 0 <= index < len(items):
    print(items[index])
else:
    print("Index is out of range")

What this does:

len(items) gets the number of items in the list
0 <= index < len(items) checks whether the index is valid
If the index is valid, Python reads the item
If not, the program avoids the error

Output:

Use len() to make sure the index exists before reading from the list.

What this error means

A list index is a position in a list.

For example, in this list:

colors = ["red", "green", "blue"]

The indexes are:

0 → "red"
1 → "green"
2 → "blue"

Python raises IndexError: list index out of range when you ask for a position that is not in the list.

Valid list indexes

For a list with length 3:

valid positive indexes are 0, 1, and 2
valid negative indexes are -1, -2, and -3

Anything outside that range will fail.

If you need a refresher on list basics, see Python lists explained for beginners.

Simple example that causes the error

Here is a short list with 3 items:

items = ["a", "b", "c"]
print(items[3])

This will raise:

IndexError: list index out of range

Why this fails

The list has 3 items, but its valid indexes are:

So items[3] does not exist.

A common beginner mistake is thinking that a 3-item list has indexes 1, 2, and 3. In Python, list indexes start at 0.

Why it happens

This error can happen in several common situations.

Using an index that is too large

numbers = [10, 20, 30]
print(numbers[5])

Index 5 is outside the list.

Using a negative index that goes too far

Negative indexes count from the end:

-1 is the last item
-2 is the second-to-last item

But this still has limits:

numbers = [10, 20, 30]
print(numbers[-4])

This fails because the list only supports negative indexes down to -3.

Looping with the wrong `range()` stop value

This is very common:

items = ["a", "b", "c"]

for i in range(len(items) + 1):
    print(items[i])

This loop goes through:

Index 3 does not exist, so the loop crashes at the end.

If you want to understand loop boundaries better, see the Python range() function explained.

Accessing a list before checking that it has items

items = []
print(items[0])

This fails because the list is empty.

Assuming input always has enough items

Sometimes the list comes from user input, split(), a file, or an API.

text = "Alice"
parts = text.split(",")

print(parts[1])

This fails because parts only has one item.

How to fix it

There are several safe ways to fix this error.

1. Check the list length with `len()` before indexing

items = ["a", "b", "c"]
index = 1

if 0 <= index < len(items):
    print(items[index])
else:
    print("Invalid index")

This is one of the clearest fixes for beginners.

2. Use the correct loop range

If you really need indexes, use:

items = ["a", "b", "c"]

for i in range(len(items)):
    print(i, items[i])

Output:

0 a
1 b
2 c

Notice that the loop stops before len(items).

3. Prefer direct iteration when possible

In many cases, you do not need indexes at all.

items = ["a", "b", "c"]

for item in items:
    print(item)

This is often safer and simpler.

For more help, see how to loop through a list in Python.

4. Check for an empty list before using `my_list[0]`

items = []

if items:
    print(items[0])
else:
    print("The list is empty")

In Python, an empty list is treated as False, so if items: is a simple emptiness check.

5. Print the index and list length while debugging

If you are not sure what is going wrong, print both values:

items = ["a", "b", "c"]
index = 3

print("List:", items)
print("Length:", len(items))
print("Index:", index)

if 0 <= index < len(items):
    print(items[index])
else:
    print("Index is out of range")

This makes it easier to see the mismatch.

Common real examples

These are common ways beginners run into this error.

Reading the first item from an empty list

results = []

if results:
    print(results[0])
else:
    print("No results found")

Using `i + 1` near the end of a loop

items = ["a", "b", "c"]

for i in range(len(items) - 1):
    print(items[i], items[i + 1])

This works because the loop stops before i + 1 goes out of range.

Splitting text and expecting more parts than exist

name = "Alice"
parts = name.split()

if len(parts) >= 2:
    print(parts[1])
else:
    print("There is no second word")

Accessing results from a file or API without checking length

results = ["first result"]

if len(results) >= 2:
    print(results[1])
else:
    print("There is no second result")

When data comes from outside your program, always check its length first.

Debugging steps

If you are getting this error and do not know why, try these steps.

1. Print the list

print(my_list)

This shows what is actually in the list.

2. Print the list length

print(len(my_list))

This tells you the highest valid positive index: len(my_list) - 1.

3. Print the index value

print(index)

Sometimes the index is not what you expected.

4. Check whether the list is empty

if not my_list:
    print("The list is empty")

This is important before using my_list[0].

5. Confirm loop boundaries

If the error happens in a loop, check your range() carefully.

Bad:

for i in range(len(my_list) + 1):
    print(my_list[i])

Good:

for i in range(len(my_list)):
    print(my_list[i])

If you need both the index and the value, enumerate() is often a better choice.

Ways to avoid this error

These habits help prevent the problem before it happens.

Loop over items instead of indexes

items = ["a", "b", "c"]

for item in items:
    print(item)

This avoids manual index mistakes.

Use `enumerate()` when you need both index and value

items = ["a", "b", "c"]

for i, value in enumerate(items):
    print(i, value)

Output:

0 a
1 b
2 c

This is safer than managing the index yourself in many cases.

Check list length before reading fixed positions

items = ["apple", "banana"]

if len(items) > 1:
    print(items[1])

Use slicing when a missing range should not crash the program

Slicing is often safer because it does not raise IndexError when the end is too large.

items = ["a", "b", "c"]
print(items[0:5])

Output:

['a', 'b', 'c']

That can be useful when you want a range of items without crashing.

Common mistakes

These are some very common causes of this error:

Trying to access my_list[len(my_list)]
Using range(len(my_list) + 1)
Reading my_list[0] when the list is empty
Using a negative index smaller than -len(my_list)
Expecting split() to return more items than it did

Why `my_list[len(my_list)]` is wrong

If a list has length 3, the valid indexes are 0, 1, and 2.

So this is wrong:

my_list = ["a", "b", "c"]
print(my_list[len(my_list)])

Because len(my_list) is 3, and index 3 is out of range.

FAQ

Why does index 3 fail in a list with 3 items?

Because list indexes start at 0. A 3-item list has valid indexes 0, 1, and 2.

Can negative indexes cause this error?

Yes. Negative indexes work only if they stay within the list length, such as -1 or -2 for a non-empty list.

How do I check if an index is valid?

Use a condition like this before accessing the item:

if 0 <= index < len(my_list):
    print(my_list[index])

Should I use try-except for this?

You can, but for beginners it is often clearer to check the length first when missing items are expected.

For example:

items = ["a", "b", "c"]
index = 5

try:
    print(items[index])
except IndexError:
    print("Index is out of range")

This works, but if out-of-range access is a normal possibility, checking first is often easier to read.