Find the index of an item in a Python list

Use items.index(value) to get the zero-based index of the first equal item. If no item matches, it raises ValueError.

fruits = ['apple', 'banana', 'cherry', 'banana']
assert fruits.index('banana') == 1
assert fruits.index('banana', 2) == 3  # Search from index 2

An optional end argument limits the search to a slice-like interval; the returned index still refers to the original list.

Find every match or search by a condition

matches = [i for i, fruit in enumerate(fruits) if fruit == 'banana']
assert matches == [1, 3]

numbers = [3, 8, 15, 7]
first = next((i for i, number in enumerate(numbers) if number > 10), None)
assert first == 2

The default None distinguishes “not found” from a valid index. Without a default, next() raises StopIteration when there is no match.

Handle a missing value

try:
    index = fruits.index('mango')
except ValueError:
    index = None

You can check 'mango' in fruits first, but a successful search then scans for the item twice. Avoid accidentally using -1 as a missing-value sentinel to index the list: in Python, it selects the last element.

FAQ

How expensive is index()?

It performs a linear search, taking O(n) comparisons in the worst case. For repeated lookups, a dictionary can map values to positions if you define how duplicates should be handled. A set answers membership, not position. Binary search requires sorted data and an equality check at the candidate position.

What about NumPy arrays?

For a one-dimensional array, np.where(array == value)[0] returns matching indices. With a multidimensional array, np.where returns one index array per dimension; taking only [0] is not a complete coordinate list.

See Python built-in types, Python built-in functions.