Many a time while developing web and desktop products in Python, we use nested lists and have several queries about how to find the count of unique tuples. Let us see how to get the count of unique tuples in the given list of tuples. Below are some ways to achieve the above task.
Method #1: Using Iteration
Python3
# Python code to count unique # tuples in list of listimport collections Output = collections.defaultdict(int)# List initializationInput = [[('hi', 'bye')], [('Geeks', 'forGeeks')], [('a', 'b')], [('hi', 'bye')], [('a', 'b')]]# Using iterationfor elem in Input: Output[elem[0]] += 1 # Printing outputprint(Output) |
defaultdict(<class 'int'>, {('Geeks', 'forGeeks'): 1, ('hi', 'bye'): 2, ('a', 'b'): 2})
Time complexity: O(n), where n is the total number of tuples in the list of lists.
Auxiliary space: O(m), where m is the total number of unique tuples in the list of lists.
Method #2: Using chain and Counter
Python3
# Python code to count unique # tuples in list of list# Importingfrom collections import Counterfrom itertools import chain# List initializationInput = [[('hi', 'bye')], [('Geeks', 'forGeeks')], [('a', 'b')], [('hi', 'bye')], [('a', 'b')]]# Using counter and chainOutput = Counter(chain(*Input))# Printing outputprint(Output) |
Counter({('hi', 'bye'): 2, ('a', 'b'): 2, ('Geeks', 'forGeeks'): 1})
Time complexity: O(n), where n is the total number of tuples in the input list of lists.
Auxiliary space: O(n), as we are using a Counter object to store the counts of each unique tuple, which can take up to n space in the worst case.
Method #3: List Comprehension method
Python3
Input = [('hi', 'bye'),('Geeks', 'forGeeks'),('a', 'b'),('hi', 'bye'),('a', 'b')]check_ele=('a', 'b') x=[i for i in Input if i==check_ele] print("tuple ('a', 'b') occurs",len(x),"times") |
tuple ('a', 'b') occurs 2 times
Time complexity: O(n), where n is the length of the input list.
Auxiliary space: O(k), where k is the number of occurrences of the check element in the input list.
Method #4: Using enumerate function
Python3
Input = [('hi', 'bye'),('Geeks', 'forGeeks'),('a', 'b'),('hi', 'bye'),('a', 'b')]check_ele=('a', 'b') x=[i for a,i in enumerate(Input) if i==check_ele] print("tuple ('a', 'b') occurs",len(x),"times") |
tuple ('a', 'b') occurs 2 times
Time complexity: O(n), where n is the length of the input list ‘Input’.
Auxiliary space: O(m), where m is the number of occurrences of the tuple ‘check_ele’ in the input list ‘Input’
Method #5: Using lambda function
Python3
Input = [('hi', 'bye'),('Geeks', 'forGeeks'),('a', 'b'),('hi', 'bye'),('a', 'b')]check_ele=('a', 'b') x=list(filter(lambda i:(i==check_ele),Input))print("tuple ('a', 'b') occurs",len(x),"times") |
tuple ('a', 'b') occurs 2 times
Time complexity: O(n), where n is the number of tuples in the Input list.
Auxiliary space: O(1) because it uses only a constant amount of extra space to store the filtered list and the length of that list.
Method #6: Using only Counter function
Python3
from collections import CounterInput = [('hi', 'bye'),('Geeks', 'forGeeks'),('a', 'b'),('hi', 'bye'),('a', 'b')]check_ele=('a', 'b')x=Counter(Input) print("tuple ('a', 'b') occurs",x[check_ele],"times") |
tuple ('a', 'b') occurs 2 times
Time complexity: O(n), where n is the number of tuples in the Input list.
Auxiliary space: O(n), where n is the number of tuples in the Input list.
Method #7: Using countof function
Python3
import operator as opInput = [('hi', 'bye'),('Geeks', 'forGeeks'),('a', 'b'),('hi', 'bye'),('a', 'b')]check_ele=('a', 'b') print(op.countOf(Input,check_ele)) |
2
Time complexity: O(n), where n is the length of the input list.
Auxiliary space: O(n), because the list comprehension creates a new list of all occurrences of the check_ele tuple.
Method #8: Using dict() method
Creating empty dictionary and iterating every element in list. Checking if element in dictionary increment value of key i by 1 else assign the key to element i with 1. finally, the dictionary will have keys as elements and their values as count of element in list.
Python3
#Initializing tuples in a listInput = [('hi', 'bye'),('Geeks', 'forGeeks'),('a', 'b'),('hi', 'bye'),('a', 'b')]#creating empty dictionary count_tuples=dict()#using for loop to iterate every value in Inputfor i in Input: if i not in count_tuples: #checking if element i present in dictionary or not count_tuples[i]=1 else: count_tuples[i]+=1#printing dictionary of elements and their countprint(count_tuples) |
{('hi', 'bye'): 2, ('Geeks', 'forGeeks'): 1, ('a', 'b'): 2}
Time complexity: O(n), where n is the length of the input list. The for loop iterates over every element in the input list exactly once.
Auxiliary space: O(m), where m is the number of unique tuples in the input list. In the worst case, when all the tuples are unique, the dictionary will contain m key-value pairs.
Method #9: Using map() function and tuple() constructor
This method involves mapping each list of tuples to a tuple of tuples and then counting the occurrences of each tuple in the list.
Python3
# Python code to count unique # tuples in list of list# List initializationInput = [[('hi', 'bye')], [('Geeks', 'forGeeks')], [('a', 'b')], [('hi', 'bye')], [('a', 'b')]]# Using map() function and tuple() constructorOutput = {}for elem in Input: t = tuple(map(tuple, elem)) Output[t] = Output.get(t, 0) + 1# Printing outputprint(Output) |
{(('hi', 'bye'),): 2, (('Geeks', 'forGeeks'),): 1, (('a', 'b'),): 2}
Time Complexity: O(nk), where n is the number of elements in the input list and k is the maximum number of tuples in a sublist.
Auxiliary Space: O(n), where n is the number of elements in the input list.
Method #10: Using set() and count()
- Initializes a list of tuples input_list containing five tuples.
- Converts the list to a set unique_set to remove duplicate tuples.
- Creates a dictionary output_dict where each key is a unique tuple from input_list, and each value is the count of that tuple in input_list.
- Prints the dictionary output_dict.
Python3
# Initializing tuples in a listinput_list = [('hi', 'bye'),('Geeks', 'forGeeks'),('a', 'b'),('hi', 'bye'),('a', 'b')]# Converting list to set to remove duplicatesunique_set = set(input_list)# Counting occurrences of each unique tupleoutput_dict = {elem: input_list.count(elem) for elem in unique_set}# Printing outputprint(output_dict) |
{('a', 'b'): 2, ('hi', 'bye'): 2, ('Geeks', 'forGeeks'): 1}
The time complexity is O(n^2)
The auxiliary space is also O(n^2)
