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Python | Program to count duplicates in a list of tuples

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  • Last Updated : 29 Dec, 2022
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Given a list of tuples, write a Python program to check if an element of the list has duplicates. If duplicates exist, print the number of occurrences of each duplicate tuple, otherwise print “No Duplicates”.

Examples:

Input  : [('a', 'e'), ('b', 'x'), ('b', 'x'), ('a', 'e'), ('b', 'x')] 
Output : ('a', 'e') - 2 ('b', 'x') - 3 

Input  : [(0, 5), (6, 9), (0, 8)] 
Output : No Duplicates

Let’s see the various ways we can count duplicates in a list of tuples. 

Approach #1: Naive Approach This approach uses two loops to traverse the list of elements and check if the first element and second element of each element match any other tuple. 

Python3




# Python3 code to convert tuple
# into string
 
 
def count(listOfTuple):
 
    flag = False
 
    # To append Duplicate elements in list
    coll_list = []
    coll_cnt = 0
    for t in listOfTuple:
 
        # To check if Duplicate exist
        if t in coll_list:
            flag = True
            continue
 
        else:
            coll_cnt = 0
            for b in listOfTuple:
                if b[0] == t[0] and b[1] == t[1]:
                    coll_cnt = coll_cnt + 1
 
            # To print count if Duplicate of element exist
            if(coll_cnt & gt
               1):
                print(t, & quot
                      - ", coll_cnt)
            coll_list.append(t)
 
    if flag == False:
        print(& quot
               No Duplicates & quot
               )
 
 
# Driver code
print(& quot
       Test Case 1: & quot
       )
listOfTuple = [('a', 'e'), ('b', 'x'), ('b', 'x'),
               ('a', 'e'), ('b', 'x')]
 
count(listOfTuple)
 
print(& quot
       Test Case 2: & quot
       )
listOfTuple = [(0, 5), (6, 9), (0, 8)]
count(listOfTuple)

Output:

Test Case 1:
('a', 'e') - 2
('b', 'x') - 3

Test Case 2:
No Duplicates

Time complexity – O(n)2   

Approach #2: Using Counter is a container included in the collections module. It is an unordered collection where elements and their respective count are stored as dictionary. 

Python3




# Python3 code to convert tuple
# into string
import collections
 
 
def count(listOfTuple):
 
    flag = False
    val = collections.Counter(listOfTuple)
    uniqueList = list(set(listOfTuple))
 
    for i in uniqueList:
        if val[i] & gt
        = 2:
            flag = True
            print(i, & quot
                  - ", val[i])
 
    if flag == False:
        print(& quot
               Duplicate doesn't exist & quot
               )
 
 
# Driver code
listOfTuple = [('a', 'e'), ('b', 'x'), ('b', 'x'),
               ('a', 'e'), ('b', 'x')]
count(listOfTuple)

Output:

('b', 'x') - 3
('a', 'e') - 2

Time complexity – O(n)   

Approach #3: Using another dict You can make a dictionary, say count_map, and store the count of each tuple as the value. 

Python3




# Python3 code to convert tuple
# into string
 
 
def count(listOfTuple):
 
    count_map = {}
    for i in listOfTuple:
        count_map[i] = count_map.get(i, 0) + 1
    print(count_map)
 
 
# Driver code
print( & quot
      Test Case 1: & quot
      )
listOfTuple = [('a', 'e'), ('b', 'x'), ('b', 'x'),
               ('a', 'e'), ('b', 'x')]
 
count(listOfTuple)

Output:

Test Case 1:
{('a', 'e'): 2, ('b', 'x'): 3}

Time complexity – O(n)

Approach #4:  Using collections

Another approach to count duplicates in a list of tuples is to use a Counter object from the collections module. You can pass the list of tuples to the Counter constructor and it will create a dictionary-like object that stores the count of each tuple. Then, you can iterate over the Counter object and print out the tuple and its count if the count is greater than 1.

Here is an example of how you could implement this approach:

Python3




from collections import Counter
 
def count_duplicates(lst):
    counter = Counter(lst)
    for tup, count in counter.items():
        if count > 1:
            print(f"{tup}: {count}")
    if not any(count > 1 for count in counter.values()):
        print("No Duplicates")
 
# Example usage
lst = [('a', 'e'), ('b', 'x'), ('b', 'x'), ('a', 'e'), ('b', 'x')]
count_duplicates(lst)  # prints ('a', 'e'): 2, ('b', 'x'): 3
 
lst = [(0, 5), (6, 9), (0, 8)]
count_duplicates(lst)  # prints No Duplicates
#This code is contributed by Edula Vinay Kumar Reddy

Output

('a', 'e'): 2
('b', 'x'): 3
No Duplicates

This approach has a time complexity of O(n), where n is the length of the input list, because it involves one pass through the list of tuples to create the Counter object and another pass through the Counter object to print the output. The space complexity is also O(n), because the Counter object stores a count for each tuple in the input list.


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