In Python, we can bind structural information in form of tuples and then can retrieve the same. But sometimes we require the information of tuple corresponding to maximum value of other tuple indexes. This functionality has many applications such as ranking. Let’s discuss certain ways in which this can be achieved.
Method #1 : Using max() + operator.itemgetter() We can get the maximum of corresponding tuple index from a list using the key itemgetter index provided and then mention the index information required using index specification at the end.
Python3
# Python3 code to demonstrate# to get tuple info. of maximum value tuple# using max() + itemgetter()from operator import itemgetter# initializing listtest_list = [('Rash', 143), ('Manjeet', 200), ('Varsha', 100)]# printing original listprint ("Original list : " + str(test_list))# using max() + itemgetter()# to get tuple info. of maximum value tupleres = max(test_list, key = itemgetter(1))[0]# printing resultprint ("The name with maximum score is : " + res) |
Original list : [('Rash', 143), ('Manjeet', 200), ('Varsha', 100)]
The name with maximum score is : Manjeet
Time Complexity: O(n), where n is the length of the input list. This is because we’re using max() + operator.itemgetter() which has a time complexity of O(n) in the worst case.
Auxiliary Space: O(n), as we’re using additional space res other than the input list itself with the same size of input list.
Method #2 : Using max() + lambda This method is almost similar to the method discussed above, just the difference is the specification and processing of target tuple index for maximum is done by lambda function. This improved readability of code.
Python3
# Python3 code to demonstrate# to get tuple info. of maximum value tuple# using max() + lambda# initializing listtest_list = [('Rash', 143), ('Manjeet', 200), ('Varsha', 100)]# printing original listprint ("Original list : " + str(test_list))# using max() + lambda# to get tuple info. of maximum value tupleres = max(test_list, key = lambda i : i[1])[0]# printing resultprint ("The name with maximum score is : " + res) |
Original list : [('Rash', 143), ('Manjeet', 200), ('Varsha', 100)]
The name with maximum score is : Manjeet
Method #3 : Using sorted() + lambda The task performed by the max() in the above two methods can be done using reverse sorting and printing the first element. lambda function performs the task similar to above said methods.
Python3
# Python3 code to demonstrate# to get tuple info. of maximum value tuple# using sorted() + lambda# initializing listtest_list = [('Rash', 143), ('Manjeet', 200), ('Varsha', 100)]# printing original listprint ("Original list : " + str(test_list))# using sorted() + lambda# to get tuple info. of maximum value tupleres = sorted(test_list, key = lambda i: i[1], reverse = True)[0][0]# printing resultprint ("The name with maximum score is : " + res) |
Original list : [('Rash', 143), ('Manjeet', 200), ('Varsha', 100)]
The name with maximum score is : Manjeet
Method #4 : Using heapq module
One approach is using the heapq module. The heapq module provides an implementation of the heap queue algorithm, also known as the priority queue algorithm.
heapq is a python module that provides an implementation of the heap queue algorithm, also known as the priority queue algorithm. It allows you to efficiently find and extract the largest or smallest elements from a list, as well as insert new elements into the list in the correct order.
The nlargest function from heapq allows you to find the n largest elements from a list. It returns a list of the n largest elements in descending order.
Here is an example of how to use the heapq module to get the first element with the maximum value in a list of tuples:
Python3
import heapqtest_list = [('Rash', 143), ('Manjeet', 200), ('Varsha', 100)]# Negate the values in the tuples to turn them into minimums# and then use heapq.nlargest to get the smallest elementres = heapq.nlargest(1, [(v, k) for k, v in test_list])[0][1]print("The name with the maximum score is:", res)#This code is contributed by Edula Vinay Kumar Reddy |
The name with the maximum score is: Manjeet
This approach has a time complexity of O(n log n) and a space complexity of O(n), where n is the number of tuples in the list.
