IndexError: pop from empty list in Python [Solved]

# IndexError: pop from empty list in Python

The Python "IndexError: pop from empty list" occurs when we call the pop() method on an empty list.

To solve the error, use an if statement to check if the list is truthy before using the pop() method, or check if the list's length is greater than 0.

Here is an example of how the error occurs.

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my_list = []

# ⛔️ IndexError: pop from empty list
result = my_list.pop()

We called the pop() method on an empty list which caused the error.

# Checking if the list is truthy before calling pop()

One way to avoid the error is to use an if statement to check if the list is truthy before using pop().

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my_list = []

if my_list:
    result = my_list.pop()
    print(result)
else:
    print('The list is empty')

You can remove the else block if you don't need it.

All values that are not truthy are considered falsy. The falsy values in Python are:

• constants defined to be falsy: None and False.
• 0 (zero) of any numeric type
• empty sequences and collections: "" (empty string), () (empty tuple), [] (empty list), {} (empty dictionary), set() (empty set), range(0) (empty range).

Empty lists are falsy and lists that contain at least 1 item are truthy.

# Checking for the list's length before calling pop()

Alternatively, you can check for the list's length to be more explicit.

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my_list = []

if len(my_list) > 0:
    result = my_list.pop()
    print(result)
else:
    print('The list is empty')

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).

If the list has a length greater than 0, we can safely call the pop() method.

# Using a try/except statement to handle the error

You can also use a try/except block to handle the error.

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my_list = []

try:
    result = my_list.pop()
    print(result)
except IndexError:
    # 👇️ this runs
    print('The list is empty')

If calling the pop() method on the list raises an IndexError, the except block is run where we can handle the error or use the pass keyword to ignore it.

Here is an example that uses a pass statement.

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my_list = []

try:
    result = my_list.pop()
    print(result)
except IndexError:
    pass

The pass statement does nothing and is used when a statement is required syntactically but the program requires no action.

# How the list.pop() method works in Python

The list.pop method removes the item at the given position in the list and returns it.

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my_list = ['bobby', 'hadz', 'com']

my_list.pop(1)

print(my_list)  # 👉️ ['bobby', 'com']

The code sample removes and returns the list item at index 1.

If no index is specified, the pop() method removes and returns the last item in the list.

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my_list = ['bobby', 'hadz', 'com']

my_list.pop()

print(my_list)  # 👉️ ['bobby', 'hadz']

# Additional Resources

You can learn more about the related topics by checking out the following tutorials:

• IndexError: index 0 is out of bounds for axis 0 with size 0
• IndexError: invalid index to scalar variable in Python
• IndexError: list assignment index out of range in Python
• Only integers, slices (:), ellipsis (...), numpy.newaxis (None) and integer or boolean arrays are valid indices