Python: IndexError: list index out of range

Python stops with:

IndexError: list index out of range

This one is refreshingly literal: you asked a list for an item at a position it doesn't have. A list of 3 items has positions 0, 1, 2 — ask for [3] and there's nothing there, so Python raises rather than guess. The fix is almost always one of three small mistakes.

Cause 1: Off-by-one (the classic)

Indexing starts at 0, so the last valid index of a list of length n is n - 1, not n. Reaching for list[len(list)] is always one too far:

items = ['a', 'b', 'c']
items[3]        # IndexError — valid indexes are 0, 1, 2
items[-1]       # 'c'  ← the safe way to get the last item

Cause 2: The list is empty

If the list has nothing in it, every index is out of range — including [0]. This bites when data didn't load, a filter removed everything, or a file was blank:

results = [r for r in data if r.active]
first = results[0]      # IndexError if nothing was active

Cause 3: A hard-coded position that isn't guaranteed

Splitting a line and grabbing parts[1] assumes there's always a second piece. A malformed row breaks it:

name, age = line.split(',')   # blows up if a line has no comma

The error is telling the exact truth: that slot doesn't exist. Don't reach for a position without knowing the list is long enough to have it.

The fixes

Check length or use safe access before indexing:

if items:                 # not empty
    print(items[0])

if len(parts) >= 2:       # enough pieces
    print(parts[1])

last = items[-1] if items else None   # safe "last item"

And when you just want to walk everything, loop instead of indexing by number:

for item in items:        # no index math, no IndexError
    print(item)

The checklist

1. You asked for a position the list doesn't have — indexes are 0..len-1.
2. Off-by-one? Use [-1] for the last item.
3. Might be empty? Guard with if items: before [0].
4. Prefer for item in items: over indexing by number where you can.