Python - Reverse Row sort in Lists of List

We are given a list of lists where each inner list represents a row and our task is to sort each row in descending order while keeping the overall structure intact. For example, given a = [[5, 2, 8], [9, 1, 4], [6, 7, 3]], after sorting each row in reverse order, the result will be [[8, 5, 2], [9, 4, 1], [7, 6, 3]].

Using NumPy

NumPy is optimized for array operations, making sorting faster than Python’s built-in functions when handling large datasets. np.sort() sorts each row, and [:, ::-1] reverses the order.

import numpy as np

a = np.array([[5, 2, 8], [9, 1, 4], [6, 7, 3]])

# Using NumPy’s sort()
res = np.sort(a)[:, ::-1]

print(res)

[[8 5 2]
 [9 4 1]
 [7 6 3]]

Explanation: np.sort(a) sorts all rows in ascending order and [:, ::-1] reverses each row to descending order.

Using heapq

heapq module in Python provides an implementation of the heap queue (priority queue) algorithm, which allows efficient extraction of the smallest or largest elements from a collection, heapq.nlargest(k, iterable) returns the k largest elements from the given iterable in descending order.

import heapq

a = [[5, 2, 8], [9, 1, 4], [6, 7, 3]]

# Using heapq.nlargest()
res = [heapq.nlargest(len(row), row) for row in a]

print(res)

[[8, 5, 2], [9, 4, 1], [7, 6, 3]]

Explanation: heapq.nlargest(len(row),row) reverses every row in "a" and the final result in saved in the list res.

Using map() + sorted()

Another approach to solving this problem is using the map() function, which can be used to apply sorted(reverse=True) to each sublist, updating the existing list with rows sorted in descending order.

a = [[4, 1, 6], [7, 8], [4, 10, 8]]

res = list(map(lambda x: sorted(x, reverse=True), a))

print(str(res))

[[6, 4, 1], [8, 7], [10, 8, 4]]

Explanation: map() applies sorted(x, reverse=True) to each sublist in 'a', sorting them in descending order and then list() method converts it to a list.

Using list comprehension + sorted()

This is yet another way in which we can pack the logic in one line using list comprehension and sorted() function with reverse = True to provide a compact alternative.

a = [[4, 1, 6], [7, 8], [4, 10, 8]]

res = [sorted(row, reverse=True) for row in a]

print(str(res))

[[6, 4, 1], [8, 7], [10, 8, 4]]

Explanation: sorted(row, reverse=True) sorts each sublist in descending order, and list comprehension applies this to all rows, storing the result in res without modifying the original list.

Using loop + sort()

We can solve this problem by using a loop to iterate through each row. The sort() function along with reverse=True can be used to sort the rows in descending order.

a = [[4, 1, 6], [7, 8], [4, 10, 8]]

for row in a:
    row.sort(reverse=True)

print(str(a))

[[6, 4, 1], [8, 7], [10, 8, 4]]

Explanation: sort(reverse=True) method sorts each sublist in descending order based on its elements, modifying the original list in place.