Python – Sort Matrix by Palindrome count

When it is required to sort a matrix based on palindrome count, a method is defined that takes a list as parameter. It uses list comprehension and the join() method to iterate and check if an element is a palindrome or not. Based on this count, the matrix rows are sorted and displayed.

What is a Palindrome?

A palindrome is a word that reads the same forwards and backwards, like "level", "noon", or "abcba".

Example

Below is a demonstration of sorting a matrix by palindrome count ?

def get_palindrome_count(row):
    return len([element for element in row if ''.join(list(reversed(element))) == element])

my_list = [["abcba", "hdgfue", "abc"], ["peep"], ["py", "is", "best"], ["sees", "level", "non", "noon"]]

print("The list is:")
print(my_list)

my_list.sort(key=get_palindrome_count)

print("The resultant list is:")
print(my_list)

The list is:
[['abcba', 'hdgfue', 'abc'], ['peep'], ['py', 'is', 'best'], ['sees', 'level', 'non', 'noon']]
The resultant list is:
[['py', 'is', 'best'], ['abcba', 'hdgfue', 'abc'], ['peep'], ['sees', 'level', 'non', 'noon']]

How It Works

Let's break down the palindrome counting process ?

def get_palindrome_count(row):
    palindromes = []
    for element in row:
        reversed_element = ''.join(list(reversed(element)))
        if reversed_element == element:
            palindromes.append(element)
            print(f"'{element}' is a palindrome")
        else:
            print(f"'{element}' is not a palindrome")
    return len(palindromes)

# Test with one row
test_row = ["sees", "level", "non", "noon"]
count = get_palindrome_count(test_row)
print(f"Total palindromes: {count}")

'sees' is a palindrome
'level' is a palindrome
'non' is a palindrome
'noon' is a palindrome
Total palindromes: 4

Sorting Process

The matrix is sorted based on palindrome count in each row ?

• Row 1: ["abcba", "hdgfue", "abc"] ? 1 palindrome ("abcba")

• Row 2: ["peep"] ? 1 palindrome ("peep")

• Row 3: ["py", "is", "best"] ? 0 palindromes

• Row 4: ["sees", "level", "non", "noon"] ? 4 palindromes

After sorting: 0, 1, 1, 4 palindromes respectively.

Alternative Approach

Here's a more readable version using a simpler palindrome check ?

def is_palindrome(word):
    return word == word[::-1]

def count_palindromes(row):
    return sum(1 for word in row if is_palindrome(word))

matrix = [["abcba", "hdgfue", "abc"], ["peep"], ["py", "is", "best"], ["sees", "level", "non", "noon"]]

print("Original matrix:")
for i, row in enumerate(matrix):
    print(f"Row {i+1}: {row} ? {count_palindromes(row)} palindromes")

matrix.sort(key=count_palindromes)

print("\nSorted matrix:")
for i, row in enumerate(matrix):
    print(f"Row {i+1}: {row} ? {count_palindromes(row)} palindromes")

Original matrix:
Row 1: ['abcba', 'hdgfue', 'abc'] ? 1 palindromes
Row 2: ['peep'] ? 1 palindromes
Row 3: ['py', 'is', 'best'] ? 0 palindromes
Row 4: ['sees', 'level', 'non', 'noon'] ? 4 palindromes

Sorted matrix:
Row 1: ['py', 'is', 'best'] ? 0 palindromes
Row 2: ['abcba', 'hdgfue', 'abc'] ? 1 palindromes
Row 3: ['peep'] ? 1 palindromes
Row 4: ['sees', 'level', 'non', 'noon'] ? 4 palindromes

Conclusion

Use sort() with a custom key function to sort matrix rows by palindrome count. The [::-1] slice notation provides a simpler way to check palindromes than using reversed() and join().