How to Find the Index of an Element in a List of Lists?

Problem Formulation

How to find the row and column index of the element x in the list of lists lst?

If the element does not occur in a list, the return value should be the tuple (-1, -1). If the element exists multiple times, the return value should be the (row, column) index of the first occurrence.

Here are three examples that demonstrate how your program should work under three important cases.

Example 1: Element Exists

Input: 
[[1, 2, 3],
 [4, 5, 6]]
x = 5

Output:
(1, 1)

Example 2: Element Doesn’t Exist

Input: 
[[1, 2, 3],
 [4, 5, 6]]
x = 0

Output:
(-1, -1)

Example 3: Element Exists Multiple Times

Input: 
[['Alice', 'Bob'],
 ['Carl', 'Dave', 'Emil'],
 ['Emil', 'Emil']]
x = 'Emil'

Output:
[1, 3]

Let’s dive into the solutions next!

Method 1: Basic Python For Loop & enumerate()

The simplest and most Pythonic way that finds the row and column indices in a general list of lists, is to use a nested for loop and the built-in enumerate() function to iterate over the elements and indices at the same time.

Here’s the code solution:

def find_element(x, lst):
    for i, row in enumerate(lst):
        for j, element in enumerate(row):
            if element == x:
                return (i, j)
    return (-1, -1)
  • The outer for loop iterates over the inner lists and their “row” indices using enumerate(). If you need a refresher on enumerate, check out my in-depth tutorial on the Finxter blog and watch the explainer video at the end of this section.
  • The inner loop iterates over each element in a given inner list, along with its “column” index.
  • As soon as you’ve found the element, return the tuple of the row and column indices (i, j).

Let’s run our three test cases against it!

# Test Case 1: Element Exists
lst = [[1, 2, 3],
       [4, 5, 6]]
x = 5
print(find_element(x, lst))

# Test Case 2: Element Doesn't Exist
lst = [[1, 2, 3],
       [4, 5, 6]]
x = 0
print(find_element(x, lst))

# Test Case 3: Element Exists Multiple Times
lst = [['Alice', 'Bob'],
       ['Carl', 'Dave', 'Emil'],
       ['Emil', 'Emil']]
x = 'Emil'
print(find_element(x, lst))

The output is the expected:

(1, 1)
(-1, -1)
(1, 2)

Before we dive into the next solution, feel free to find an in-depth explanation of the enumerate() function here:

Method 2: enumerate() and list.index()

An alternative way to accomplish the same task is as follows.

To find the (row, column) index pair of an element in a list of lists, iterate over the rows and their indices using the enumerate() function and use the row.index(x) method to determine the index of element x in the row.

A bit additional code is needed to make sure that if the element is not found in a given row, the raised Error is properly handled:

def find_element(x, lst):
    for i, row in enumerate(lst):
        try:
            return (i, row.index(x))
        except:
            pass
    return (-1, -1)

The try/except code block handles the cases where the element doesn’t exist in the row. Alternatively, you could use a simple check like so:

def find_element(x, lst):
    for i, row in enumerate(lst):
        if x in row:
            return (i, row.index(x))
    return (-1, -1)

This is more readable and more concise but it does a bit of additional work: if the element does exist, it searches it twice in the list, once for the membership operation and once for the row.index(x) method.

Method 3: A One-Liner Solution

if you enjoy Python one-liners like I do, you’ll love this one:

find_element = lambda x, lst: [(i, row.index(x)) for i, row in enumerate(lst) if x in row]
  • You create a variable find_element and assign a dynamic function object to it, created using the lambda keyword.
  • The function takes two arguments x and lst. The first is the element to be searched in the list of lists lst.
  • It returns a list of all found occurrences using a list comprehension statement. If you need a refresher on list comprehension, check out my detailed guide on the Finxter blog and watch the explainer video at the end of this section.
  • In the list comprehension statement, we iterate over all rows and their indices and use the row.index(x) method to check the index of the first occurrence of x in row if it exists.

Note that the output is now a list of tuples—each representing one found occurrence. However, if you strictly need the formatting requirements defined in the problem formulation, you can slightly modify it like so:

def find_element(x, lst):
    res = [(i, row.index(x)) for i, row in enumerate(lst) if x in row]
    return res[0] if res else (-1, -1)

Here’s the promised explainer on list comprehension—a critical Python feature:

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