Given a list of tuples with word as first element and its frequency as second element, the task is to find top k frequent element.
Below are some ways to above achieve the above task.
Method #1: Using defaultdict
# Python code to find top 'k' frequent element # Importing import collections from operator import itemgetter from itertools import chain # Input list initialization Input =[[('Name', 151)], [('ACe', 400)], [('TURN', 210)], [('RED', 1113)], [('YELLOW', 1)]] # K initialization K = 3 # Using defaultdict to find top 'k' frequent element dict_ = collections.defaultdict(list) new_list = list(chain.from_iterable(Input)) for elem in new_list: dict_[elem[0]].append(elem[1]) res = {k: sum(v) for k, v in dict_.items()} # Using sorted Output = sorted(res.items(), key = itemgetter(1), reverse = True)[0:K] # printing output print("Initial List of tuple is", Input) print("\nTop 'K' elements are", Output) |
Initial List of tuple is [[(‘Name’, 151)], [(‘ACe’, 400)], [(‘TURN’, 210)], [(‘RED’, 1113)], [(‘YELLOW’, 1)]]
Top ‘K’ elements are [(‘RED’, 1113), (‘ACe’, 400), (‘TURN’, 210)]
Method #2: Using itertools and sorted
# Python code to find top 'k' frequent element from operator import itemgetter from itertools import chain # Input list initialization Input =[[('Name', 151)], [('ACe', 400)], [('TURN', 210)], [('RED', 1113)], [('YELLOW', 1)]] # k initialization K = 3 # Finding top 'k' frequent element # without using collection Output = sorted(list(chain.from_iterable(Input)), key = itemgetter(1), reverse = True)[0:K] # Printing Output print("Initial List of tuple is", Input) print("\nTop 'K' elements are", Output) |
Initial List of tuple is [[(‘Name’, 151)], [(‘ACe’, 400)], [(‘TURN’, 210)], [(‘RED’, 1113)], [(‘YELLOW’, 1)]]
Top ‘K’ elements are [(‘RED’, 1113), (‘ACe’, 400), (‘TURN’, 210)]
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