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Python | Remove elements with a frequency of no more than K

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Method # 1: Using list comprehension + count()
The idea applied here is to create a new list using list comprehension and insert only those elements that occur more than K times. The counting operation is performed using the counting function.

# Python3 demo code
# remove items less and equal to K
# using list comprehension + count ()

 
# initializing list

test_list = [ 1 , 4 , 3 , 2 , 3 , 3 , 2 , 2 , 2 , 1 ]

 
# print original list

print ( "The original list:" + str (test_list ))

 
# initializing K

K = 2

 
# using a list comprehension + count ()
# remove items less than K

res = [i for i in test_list if test_list.count (i)" K]

 
# print result

print ( "The list removing elements less than and equal K :" + str (res))

Output:

 The original list: [1, 4, 3, 2, 3, 3, 2, 2, 2, 1] The list removing elements less than and equal K: [3, 2, 3, 3, 2, 2, 2] 

Method # 2: Using Counter () + List Comprehension
This problem can be effectively solved by using the Counter function, which pre-calculates the number of each item in the list, so the solution is it takes less time to save or reject a specific element.

< code class = "comments"> # Python3 demo code
# remove items less than and equal to K
# using Counter () + list comprehension

from collections import Counter

 
# initializing list

test_list = [ 1 , 4 , 3 , 2 , 3 , 3 , 2 , 2  , 2 , 1 ]

 
# prints the original list

print ( "The original list : " + str (test_list))

 
# initialization K

K = 2

  
# using Counter () + list comprehension
# remove items less than K

freq = Counter (test_list)

res = [ele for ele in test_list if freq [ele]" K]

 
# print result

print ( "The list removing elements less than and equal K :" + str (res))

Output:

 The original list: [1, 4, 3, 2, 3, 3, 2, 2, 2, 1] The list removing elements less than and equal K: [3, 2, 3, 3, 2, 2, 2] 

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