An IndexError occurs in Python when you try to access an element of a sequence (list, tuple, or string) using an index that is outside the valid range. Python sequences are zero-indexed, meaning the first element is at index 0 and the last is at index len(sequence) - 1. Accessing any index beyond this range raises an IndexError.
# ❌ Problem
items = [10, 20, 30]
print(items[3]) # IndexError: list index out of range (valid: 0, 1, 2)
# ✅ Solution
print(items[2]) # 30 "” last valid index
print(items[-1]) # 30 "” negative index for last elementTrying to access any index of an empty list raises an IndexError immediately. Always check if a list is non-empty before accessing its elements.
A classic off-by-one error occurs when using range(len(list)) but accidentally going one step too far, or when manually incrementing an index counter past the last valid position.
Using a hardcoded index assumes the list always has a certain number of elements. If the list is shorter than expected (e.g., from a filtered result or API response), the index will be out of range.
When using a while loop with a manual index, it is easy to forget to stop at the right boundary. Always use while i < len(list) rather than while i
results = []
first = results[0] # IndexError: list index out of rangeresults = []
# ✅ Check before accessing
if results:
first = results[0]
else:
first = None
# ✅ Or use a try/except
try:
first = results[0]
except IndexError:
first = Nonefruits = ["apple", "banana", "cherry"]
for i in range(len(fruits) + 1): # +1 causes IndexError on last iteration
print(fruits[i])fruits = ["apple", "banana", "cherry"]
# ✅ Best: iterate directly over the list
for fruit in fruits:
print(fruit)
# ✅ If you need the index, use enumerate()
for i, fruit in enumerate(fruits):
print(f"{i}: {fruit}")def get_top_scores(scores):
return scores[0], scores[1], scores[2] # IndexError if fewer than 3 scores
scores = [95, 87]
top = get_top_scores(scores) # IndexError: list index out of rangedef get_top_scores(scores, n=3):
# ✅ Use slicing "” never raises IndexError
return scores[:n]
scores = [95, 87]
top = get_top_scores(scores) # [95, 87] "” no erroritems = [1, 2, 3]
i = 0
while i <= len(items): # Should be < not <=
print(items[i]) # IndexError on last iteration when i == 3
i += 1items = [1, 2, 3]
i = 0
while i < len(items): # ✅ Strict less-than
print(items[i])
i += 1
# ✅ Even better: use a for loop
for item in items:
print(item)It means you tried to access an index that does not exist in the list. For a list of 3 elements, valid indices are 0, 1, and 2. Accessing index 3 or higher raises this error.
Use negative indexing: list[-1] returns the last element. But first check the list is not empty: if my_list: last = my_list[-1].
IndexError occurs with sequences (lists, tuples, strings) when an integer index is out of range. KeyError occurs with dictionaries when a key does not exist.
Use a for-in loop (for item in list) which automatically handles boundaries. If you need indices, use enumerate(). Avoid manual index management with while loops when possible.
Python does not have a built-in .get() for lists like dicts do. Use try/except IndexError, check the length first, or use slicing. You can also use next(iter(list[n:]), default) as a one-liner.
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