Table of Contents

## Problem Formulation

- Given a list of lists

, and**lst** - Given an element

.**x**

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 = 5Output:(1, 1)

### Example 2: Element Doesn’t Exist

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