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Python IndexError: list index out of range

The code tries to access a list index that is beyond the list's length.

Explanation

IndexError occurs when you try to access a list (or other sequence) element at an index that doesn't exist. List indices in Python start at 0, so a list of length 3 has valid indices 0, 1, and 2. Accessing index 3 or higher raises IndexError. This commonly happens when iterating beyond list bounds, when assuming a list has a certain number of elements, or when negative indexing goes beyond the list length.

Common Causes

  • Index exceeds list length
  • Empty list accessed by index
  • Off-by-one error in loops
  • Assuming list has certain elements
  • Incorrect array index calculation

Solution

Check the list length before accessing by index using len(). Use try-except IndexError blocks for safe access. Use enumerate() for index-based iteration instead of manual index management. Use list slicing which doesn't raise errors for out-of-bounds indices. For empty lists, add a check before accessing the first element. Consider using collections.deque for operations where you need to access elements from both ends.

Code Example 

# IndexError from empty list
users = []
first = users[0]  # IndexError: list index out of range

# Fix: check length first
if users:
    first = users[0]

# IndexError from out of range
items = [1, 2, 3]
fifth = items[4]  # IndexError

# Fix: check index
if len(items) > 4:
    fifth = items[4]

# Safe access function
def safe_index(lst, index, default=None):
    try:
        return lst[index]
    except IndexError:
        return default

result = safe_index(items, 4, 'N/A')

# Using enumerate for iteration (no manual index)
for i, item in enumerate(items):
    print(f'{i}: {item}')

# Slicing doesn't raise IndexError
subset = items[0:100]  # returns all available items

Error Information

Language

python

Difficulty

Beginner

Views

13

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