Open In App

Python – Maximum of Similar Keys in Tuples

Last Updated : 28 Feb, 2023
Improve
Improve
Like Article
Like
Save
Share
Report

Sometimes, while working with Python tuples, we can have a problem in which we need to perform maximum of all values of the equal keys in Tuple list. This kind of application in useful in many domains such as web development and day-day programming. Let’s discuss certain ways in which this task can be performed.

Input : test_list = [(4, 8), (4, 2), (4, 2), (4, 6)] 
Output : [(4, 8)]

Input : test_list = [(1, 8), (2, 2), (3, 6), (4, 2)] 
Output : [(1, 8), (2, 2), (3, 6), (4, 2)] 
 

Method #1 : Using max() + groupby() + lambda + loop 
The combination of above functions can be used to solve this problem. In this, we perform the task of grouping using groupby() and maximum is extracted using max(), and result is compiled using loop.

Python3




# Python3 code to demonstrate working of
# Maximum of Similar Keys in Tuples
# Using max() + groupby() + lambda + loop
from itertools import groupby
 
# initializing lists
test_list = [(4, 8), (3, 2), (2, 2), (4, 6), (3, 7), (4, 5)]
 
# printing original list
print("The original list is : " + str(test_list))
 
# Maximum of Similar Keys in Tuples
# Using max() + groupby() + lambda + loop
test_list.sort(key = lambda sub: sub[0])
temp = groupby(test_list, lambda ele: ele[0])
res = []
for key, val in temp:
    res.append((key, sum([ele[1] for ele in val])))
 
# printing result
print("Maximum grouped elements : " + str(res))


Output : 

The original list is : [(4, 8), (3, 2), (2, 2), (4, 6), (3, 7), (4, 5)]
Maximum grouped elements : [(2, 2), (3, 7), (4, 8)]

 

Time complexity: O(nlogn) where n is the number of tuples in the list, because sorting the list takes O(nlogn) time.
Auxiliary space: O(n) where n is the number of tuples in the list, because res list with n elements is created.

Method #2 : Using max() + groupby() + itemgetter() + list comprehension 
The combination of above functions can be used to solve this problem. In this, we perform similar task as above, the second element is chosen using itemgetter and list comprehension is used to compile elements and extract result. 

Python3




# Python3 code to demonstrate working of
# Maximum of Similar Keys in Tuples
# Using max() + groupby() + itemgetter() + list comprehension
from itertools import groupby
from operator import itemgetter
 
# initializing lists
test_list = [(4, 8), (3, 2), (2, 2), (4, 6), (3, 7), (4, 5)]
 
# printing original list
print("The original list is : " + str(test_list))
 
# Maximum of Similar Keys in Tuples
# Using max() + groupby() + itemgetter() + list comprehension
temp = groupby(sorted(test_list, key = itemgetter(0)), key = itemgetter(0))
res = [(key, max(map(itemgetter(1), sub))) for key, sub in temp]
 
# printing result
print("Maximum grouped elements : " + str(res))


Output : 

The original list is : [(4, 8), (3, 2), (2, 2), (4, 6), (3, 7), (4, 5)]
Maximum grouped elements : [(2, 2), (3, 7), (4, 8)]

 

Time complexity: O(nlogn) – sorting the list takes nlogn time, where n is the length of the input list. The groupby function also takes linear time, but the max and map functions inside the list comprehension are negligible compared to the other operations.
Auxiliary space: O(n) – creating the sorted list and the groupby object requires additional memory proportional to the length of the input list. However, the space used by the result list is also proportional to n, so the overall space complexity is O(n).

Method #3 : Using Counter and list comprehension: 

Algorithm:

1.Create a Counter object from the given list of tuples.
2.Convert the Counter object into a dictionary.
3.Sort the keys of the dictionary.
4.Create a list of tuples from the sorted dictionary, where each tuple contains the key and its corresponding value.

Python3




from collections import Counter
 
# initializing lists
test_list = [(4, 8), (3, 2), (2, 2), (4, 6), (3, 7), (4, 5)]
  
# printing original list
print("The original list is : " + str(test_list))
 
# using Counter to count the frequency of each key
d = dict(Counter(dict(test_list)))
 
# creating a list of tuples sorted by key
res = [(k, d[k]) for k in sorted(d)]
 
# printing result
print("Maximum grouped elements : " + str(res))
 
 
#This code is contributed by Jyothi pinjala.


Output

The original list is : [(4, 8), (3, 2), (2, 2), (4, 6), (3, 7), (4, 5)]
Maximum grouped elements : [(2, 2), (3, 7), (4, 5)]

Time complexity: O(nlogn) (due to sorting of dictionary keys)
Auxiliary Space: O(n) (to store the Counter and dictionary objects)



Like Article
Suggest improvement
Previous
Next
Share your thoughts in the comments

Similar Reads