Sometimes, while working with data, we can have a problem in which we need to get the maximum of elements filtered by the Kth element of record. This has a very important utility in web development domain. Let’s discuss certain ways in which this task can be performed.
Method #1 : Using filter() + lambda + set() + list comprehension The combination of above functions can be used to perform this particular function. In this, we first filter the top N elements from Kth index and then apply these values to the list and return the result.
Python3
# Python3 code to demonstrate working of # Top N pairs by Kth element from list # Using filter() + lambda + set() + list comprehension # initialize list test_list = [('gfg', 4, 'good'), ('gfg', 2, 'better'), ('gfg', 1, 'best'), ('gfg', 3, 'neveropen')] # printing original list print("The original list is : " + str(test_list)) # initialize N N = 3# initialize K K = 1# Top N pairs by Kth element from list # Using filter() + lambda + set() + list comprehension temp = set(list({sub[K] for sub in test_list})[-N:]) res = list(filter(lambda sub: sub[K] in temp, test_list)) # printing result print("Top N elements of Kth index are : " + str(res)) |
The original list is : [('gfg', 4, 'good'), ('gfg', 2, 'better'), ('gfg', 1, 'best'), ('gfg', 3, 'neveropen')]
Top N elements of Kth index are : [('gfg', 4, 'good'), ('gfg', 2, 'better'), ('gfg', 3, 'neveropen')]
Time Complexity: O(n*nlogn) where n is the number of elements in the list “test_list”. filter() + lambda + set() + list comprehension performs n*nlogn number of operations.
Auxiliary Space: O(n), extra space is required where n is the number of elements in the list
Method #2 : Using groupby() + sorted() + loop This task can also be performed using above functionalities. In this, we first group the top N elements together and then limit by N while constructing the result list.
Python3
# Python3 code to demonstrate working of # Top N pairs by Kth element from list # Using groupby() + sorted() + loop import itertools # initialize list test_list = [('gfg', 4, 'good'), ('gfg', 2, 'better'), ('gfg', 1, 'best'), ('gfg', 3, 'neveropen')] # printing original list print("The original list is : " + str(test_list)) # initialize N N = 3# initialize K K = 1# Top N pairs by Kth element from list # Using groupby() + sorted() + loop res = [] temp = itertools.groupby(sorted(test_list, key = lambda sub : sub[K], reverse = True), key = lambda sub : sub[K]) for i in range(N): res.extend(list(next(temp)[K])) # printing result print("Top N elements of Kth index are : " + str(res)) |
The original list is : [('gfg', 4, 'good'), ('gfg', 2, 'better'), ('gfg', 1, 'best'), ('gfg', 3, 'neveropen')]
Top N elements of Kth index are : [('gfg', 4, 'good'), ('gfg', 3, 'neveropen'), ('gfg', 2, 'better')]
Time Complexity: O(n*nlogn), where n is the length of the list test_list
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the res list
Method #3: Using heapq.nlargest()
This approach uses the heapq.nlargest() function to get the top N elements from the list based on the Kth element. It is more efficient than the above two methods as it uses a heap data structure to find the top N elements in O(nlogk) time complexity.
Python3
import heapq# initialize list test_list = [('gfg', 4, 'good'), ('gfg', 2, 'better'), ('gfg', 1, 'best'), ('gfg', 3, 'neveropen')]# printing original listprint("The original list is : " + str(test_list)) # initialize N N = 3 # initialize K K = 2# Top N pairs by Kth element from list# Using heapq.nlargest()res = heapq.nlargest(N, test_list, key=lambda x: x[K])# printing resultprint("Top N elements of Kth index are : " + str(res))#This code is contributed by Edula Vinay Kumar Reddy |
The original list is : [('gfg', 4, 'good'), ('gfg', 2, 'better'), ('gfg', 1, 'best'), ('gfg', 3, 'neveropen')]
Top N elements of Kth index are : [('gfg', 4, 'good'), ('gfg', 3, 'neveropen'), ('gfg', 2, 'better')]
The time complexity of this method is O(nlogk) and Auxiliary space is O(k).
