Find top K frequent elements from a list of tuples in Python

We have a list of tuples. In it we are required to find the top k frequent element. If k is 3 we are required to find the top three elements from the tuples inside the list.

With defaultdict

We put the the elements into a dictionary container using defaultdict. Then find out only the elements which satisfy that top k conditions.

Example

 Live Demo

import collections
from operator import itemgetter
from itertools import chain
# Input list initialization
listA = [[('Mon', 126)], [('Tue', 768)],[('Wed', 512)], [('Thu', 13)],[('Fri', 341)]]
# set K
K = 3
#Given list
print("Given list:\n",listA)
print("Check value:\n",K)
# Using defaultdict
dict_ = collections.defaultdict(list)
new_list = list(chain.from_iterable(listA))
for elem in new_list:
   dict_[elem[0]].append(elem[1])
res = {k: sum(v) for k, v in dict_.items()}
# Using sorted
res = sorted(res.items(), key=itemgetter(1),
reverse=True)[0:K]
# Output
print("Top 3 elements are:\n", res)

Output

Running the above code gives us the following result −

Given list:
[[('Mon', 126)], [('Tue', 768)], [('Wed', 512)], [('Thu', 13)], [('Fri', 341)]]
Check value:
3
Top 3 elements are:
[('Tue', 768), ('Wed', 512), ('Fri', 341)]

With sorted and itergetter

In this approach we use the itemgetter function but apply it within the sorted function by mentioning the range from 0 to K.

Example

 Live Demo

from operator import itemgetter
from itertools import chain
# Input list initialization
listA = [[('Mon', 126)], [('Tue', 768)],[('Wed', 512)], [('Thu', 13)],[('Fri', 341)]]
# set K
K = 3
#Given list
print("Given list:\n",listA)
print("Check value:\n",K)
# Using sorted
res = sorted(list(chain.from_iterable(listA)),
   key = itemgetter(1), reverse = True)[0:K]
# Output
print("Top 3 elements are:\n", res)

Output

Running the above code gives us the following result −

Given list:
[[('Mon', 126)], [('Tue', 768)], [('Wed', 512)], [('Thu', 13)], [('Fri', 341)]]
Check value:
3
Top 3 elements are:
[('Tue', 768), ('Wed', 512), ('Fri', 341)]

Pradeep Elance

Published on 04-Jun-2020 15:24:22