This concise, example-based article will walk you through 3 different ways to check whether a Python list contains an element. There’s no time to waste; let’s get started!

Table Of Contents

1 Using the “in” operator

2 Using the list.index() method

3 Using the list.count() method

Using the “in” operator

This solution uses the in operator to check if an element is in a list. It returns True if the element is found and False otherwise.

# define a list with different data types
my_list = [1, 2, 3, ['a', 'b', 'c'], {'sling': 'academy'}]

print(2 in my_list) # True
print(['a', 'b', 'c'] in my_list) # True
print({'sling': 'academy'} in my_list) # True
print('turtle' in my_list) # False

The code is concise and intuitive. It also works for any type of element, even if they are not hashable or mutable.

Using the list.index() method

This solution uses the index() method to find the index of the first occurrence of an element in a list. It returns the index as an integer if the element is found and raises a ValueError exception otherwise.

Example:

# define a list with different data types
my_list = [1, 2, 3, ['a', 'b', 'c'], {'sling': 'academy'}]

try:
    index = my_list.index(3)
    print(f"'3' found at index {index}")
except ValueError:
    print('Value not found in list')

Output:

'3' found at index 2

This approach requires handling a potential exception, but it provides information about the index of the element in the list (if found).

Using the list.count() method

Another working solution is to use the list.count() method to count the occurrences of an element in a list and check if it’s greater than zero.

Example:

my_list = [1, 2, 3, 4, 5, 3, 4, 3, 3]
element = 3

count = my_list.count(element)

if count > 0:
    print(f'Element {element} is in the list {count} times')
else:
    print(f'Element {element} is not in the list')

Output: