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IndexError: list index out of range

IndexError: list index out of range in Python means you asked for an index the list does not have. Almost always a loop bound that is one too far, or an empty list you assumed had items.

PythonPattern: Off-by-one errors

How do I fix IndexError: list index out of range?

To fix IndexError: list index out of range in Python, locate the line causing the error and check the values right before it runs. Most occurrences happen because of a loop that runs one time too many. Apply a defensive check or safe fallback, make sure collections or objects exist before access, and verify return values from upstream functions.

1. Inspect bounds & typesPrint or inspect collection sizes and verify variables are not empty or None/undefined.
2. Guard with safe accessUse optional chaining (?.), dictionary get(), or guard conditions before dereferencing.
3. Trace upstream originsFind where the value was instantiated or returned rather than only patching the symptom.

What does IndexError: list index out of range mean?

Python lists are indexed from 0, so a list of 5 items has valid indices 0 to 4. Asking for index 5 — or any index at or beyond the length — raises IndexError immediately. Unlike JavaScript, which quietly returns undefined, Python fails loudly at the exact line, which is genuinely helpful once you know what it means.

Why does IndexError: list index out of range happen?

1. A loop that runs one time too many

range(len(items) + 1) or a while loop testing <= length both step one past the end.

Breaks

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

Works

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

2. Using the length as an index

The last valid index is len - 1. len itself is always one past the end.

Breaks

last = items[len(items)]

Works

last = items[len(items) - 1]   # or items[-1]

3. The list is empty

items[0] fails on an empty list. This is common when a filter or an API call returned nothing.

Breaks

first = results[0]

Works

first = results[0] if results else None

4. Removing items while looping over them

The list shrinks as you iterate, so the index eventually points past the new end.

Breaks

for i in range(len(items)):
    if items[i] < 0:
        items.pop(i)

Works

items = [x for x in items if x >= 0]

How to find it in your own code

Print len(items) and the index right before the failing line. If the index equals the length, your bound is one too far. If the length is 0, the real bug is upstream — something returned nothing and you did not check.

Try it: fix this bug

easy

count_rises([3, 5, 4, 6]) crashes: IndexError: list index out of range.

def count_rises(prices):
    rises = 0
    for i in range(len(prices)):
        if prices[i + 1] > prices[i]:
            rises += 1
    return rises
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