- Given a list of lists
- Given an element
How to find the row and column index of the element x in the list of lists
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
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
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
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_elementand assign a dynamic function object to it, created using the
- The function takes two arguments
lst. The first is the element to be searched in the list of lists
- 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
rowif 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 if res else (-1, -1)
Here’s the promised explainer on list comprehension—a critical Python feature:
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