Solve - IndexError: tuple index out of range in Python

Solve - IndexError: tuple index out of range in Python #

The Python "IndexError: tuple index out of range" occurs when we try to access an index that doesn't exist in a tuple. Indexes are zero-based in Python, so the index of the first item in the tuple is 0, and the index of the last is -1 or len(my_tuple) - 1.

Here is an example of how the error occurs.

Copied!
my_tuple = ('a', 'b', 'c')

# ⛔️ IndexError: tuple index out of range
print(my_tuple[3])

The tuple has a length of 3. Since indexes in Python are zero-based, the first item in the tuple has an index of 0, and the last an index of 2.

If you need to get the last item in a tuple, use -1.

Copied!
my_tuple = ('a', 'b', 'c')

print(my_tuple[-1])  # 👉️ c
print(my_tuple[-2])  # 👉️ b

When the index starts with a minus, we start counting backwards from the end of the tuple.

If you need to get the length of the tuple, use the len() function.

Copied!
my_tuple = ('a', 'b', 'c')

print(len(my_tuple)) # 👉️ 3

idx = 3

if len(my_tuple) > idx:
    print(my_tuple[idx])
else:
    # 👇️ this runs
    print(f'index {idx} is out of range')

The len() function returns the length (the number of items) of an object.

The argument the function takes may be a sequence (a string, tuple, list, range or bytes) or a collection (a dictionary, set, or frozen set).

This means that you can check if the tuple's length is greater than the index you are trying to access.

If you are trying to iterate over a tuple using the range() function, use the length of the tuple.

Copied!
my_tuple = ('a', 'b', 'c')

for i in range(len(my_tuple)):
    print(i, my_tuple[i])  # 👉️ 0 a, 1 b, 2 c

However, a much better solution is to use the enumerate() function to iterate over a tuple with the index.

Copied!
my_tuple = ('a', 'b', 'c')

for index, item in enumerate(my_tuple):
    print(index, item)  # 👉️ 0 a, 1 b, 2 c

The enumerate function takes an iterable and returns an enumerate object containing tuples where the first element is the index, and the second, the item.

An alternative approach to handle the "IndexError: tuple index out of range" exception is to use a try/except block.

Copied!
my_tuple = ('a', 'b', 'c')

try:
    result = my_tuple[100]
except IndexError:
    print('index out of range')

We tried accessing the tuple item at index 100 which raised an IndexError exception.

You can handle the error or use the pass keyword in the except block.

Note that if you try to access an empty tuple at a specific index, you'd always get an IndexError.

Copied!
my_tuple = ()

print(my_tuple) # 👉️ ()
print(len(my_tuple)) # 👉️ 0

# ⛔️ IndexError: tuple index out of range
print(my_tuple[0])

You should print the tuple you are trying to access and its length to make sure the variable stores what you expect.

In case you declared a tuple by mistake, tuples are constructed in multiple ways:

• Using a pair of parenthesis () creates an empty tuple
• Using a trailing comma - a, or (a,)
• Separating items with commas - a, b or (a, b)
• Using the tuple() constructor