In python we need to handle various forms of data and one among them is list of tuples in which we may have to perform any kind of operation. This particular article discusses the ways of finding the frequency of the 1st element in list of tuple which can be extended to any index. Let’s discuss certain ways in which this can be performed.
Method #1 : Using map() + count() The map function can be used to accumulate the indices of all the tuples in a list and the task of counting the frequency can be done using the generic count function of python library.
Python3
# Python3 code to demonstrate# finding frequency in list of tuples# using map() + count()# initializing list of tuplestest_list = [('Geeks', 1), ('for', 2), ('Geeks', 3)]# printing the original listprint ("The original list is : " + str(test_list))# using map() + count()# finding frequency in list of tuples res = list(map(lambda i : i[0], test_list)).count('Geeks')# printing resultprint ("The frequency of element is : " + str(res)) |
The original list is : [('Geeks', 1), ('for', 2), ('Geeks', 3)]
The frequency of element is : 2
Time Complexity: O(n), 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 #2 : Using Counter() + list comprehension List comprehension performs the task of getting the first element of the tuples and the counting part is handled by Counter function of collection library.
Python3
# Python3 code to demonstrate# finding frequency in list of tuples# using Counter() + list comprehensionfrom collections import Counter# initializing list of tuplestest_list = [('Geeks', 1), ('for', 2), ('Geeks', 3)]# printing the original listprint ("The original list is : " + str(test_list))# using Counter() + list comprehension# finding frequency in list of tuples res = Counter(i[0] for i in test_list)# printing resultprint ("The frequency of element is : " + str(res['Geeks'])) |
The original list is : [('Geeks', 1), ('for', 2), ('Geeks', 3)]
The frequency of element is : 2
Method #3 : Dictionary comprehension and the get() method
To use dictionary comprehension and the get() method to find the frequency of the element ‘Geeks’ in the list of tuples test_list, you can do the following:
Python3
# initializing list of tuplestest_list = [('Geeks', 1), ('for', 2), ('Geeks', 3)]# create a dictionary to store the frequency of each element looping and the get() methodfrequency = {}for element in test_list: frequency[element[0]] = frequency.get(element[0], 0) + 1# print the frequency of the element 'Geeks'print(frequency['Geeks'])#This code is contributed by Edula Vinay Kumar Reddy |
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If the element is already in the dictionary, the get() method returns its current value, and this value is incremented by 1. This allows us to count the frequency of each element in the list of tuples.
Time complexity: O(n)
Auxiliary Space: O(n)
Method #4 : Using map() + operator.countOf() method
The map function can be used to accumulate the indices of all the tuples in a list and the task of counting the frequency can be done using the operator.countOf function of python library.
Python3
# Python3 code to demonstrate# finding frequency in list of tuples# using map() + operator.countOf()import operator as op# initializing list of tuplestest_list = [('Geeks', 1), ('for', 2), ('Geeks', 3)]# printing the original listprint ("The original list is : " + str(test_list))# using map() + operator.countOf()# finding frequency in list of tuples res = op.countOf(list(map(lambda i : i[0], test_list)),'Geeks')# printing resultprint ("The frequency of element is : " + str(res)) |
The original list is : [('Geeks', 1), ('for', 2), ('Geeks', 3)]
The frequency of element is : 2
Time complexity: O(n*n)
Auxiliary Space: O(1)
Method #5: Using np.unique
The np.unique function in NumPy can be used to find the unique elements in a given array and their frequencies.
Algorithm:
- Initialize a NumPy array with the first elements of each tuple in the list of tuples.
- Use np.unique with the return_counts parameter set to True to get the unique elements and their frequencies.
- Find the index of the element to get its frequency.
Python3
# Python3 code to demonstrate# finding frequency in list of tuples# using np.uniqueimport numpy as np# initializing list of tuplestest_list = [('Geeks', 1), ('for', 2), ('Geeks', 3)]# printing the original listprint("The original list is : " + str(test_list))# initializing NumPy array with first elements of tuplesarr = np.array([t[0] for t in test_list])# using np.unique to get unique elements and their frequenciesunique, counts = np.unique(arr, return_counts=True)# finding the index of the elementindex = np.where(unique == 'Geeks')[0][0]# printing resultprint("The frequency of element is : " + str(counts[index])) |
Output:
The original list is : [('Geeks', 1), ('for', 2), ('Geeks', 3)]
The frequency of element is : 2
Time Complexity: O(n log n), where n is the length of the list test_list (due to the sorting operation performed by np.unique).
Auxiliary Space: O(n), where n is the number of elements in the NumPy array arr.
METHOD 6:Using re-module.
APPROACH:
This program uses regular expressions in Python to find the frequency of a given element in a list of tuples. The program first creates a regular expression pattern that matches strings that start with the desired element followed by a word boundary. It then uses a generator expression to iterate over the list of tuples and count the number of tuples whose first element matches the pattern. Finally, it prints the result.
ALGORITHM:
1. Initialize the list of tuples and the element to be searched for.
2. Create a regular expression pattern that matches strings starting with the desired element followed by a word boundary.
3. Use a generator expression to iterate over the list of tuples and count the number of tuples whose first element matches the pattern.
4. Print the result.
Python3
import relst = [('Geeks', 1), ('for', 2), ('Geeks', 3)]element = 'Geeks'pattern = re.compile(rf"^{element}\b")count = sum(1 for tpl in lst if pattern.match(tpl[0]))print("The frequency of", element, "is", count) |
The frequency of Geeks is 2
Time complexity:
The time complexity of this program is O(n), where n is the number of tuples in the list. This is because the program needs to iterate over each tuple in the list once to count the frequency of the desired element.
Space complexity:
The space complexity of this program is O(1), as the program only uses a constant amount of additional memory to store the regular expression pattern, the count variable, and the element to be searched for. The memory used by the list of tuples is not counted, as it is part of the input.
