Python – Sort rows by Frequency of K
Last Updated :
10 Mar, 2023
Given a Matrix, the task is to write a Python program to perform sorting on rows depending on the frequency of K.
Input : test_list = [[10, 2, 3, 2, 3], [5, 5, 4, 7, 7, 4], [1, 2], [1, 1, 2, 2, 2]], K = 2
Output : [[5, 5, 4, 7, 7, 4], [1, 2], [10, 2, 3, 2, 3], [1, 1, 2, 2, 2]]
Explanation : 0 < 1 < 2 < 3, count of K in Matrix order.
Input : test_list = [[5, 5, 4, 7, 7, 4], [1, 2], [1, 1, 2, 2, 2]], K = 2
Output : [[5, 5, 4, 7, 7, 4], [1, 2], [1, 1, 2, 2, 2]]
Explanation : 0 < 1 < 3, count of K in Matrix order.
Method #1 : Using sort() + count()
In this, we perform the task of in-place sorting using sort(), capturing frequency is done using count().
Python3
def get_Kfreq(row):
return row.count(K)
test_list = [[ 10 , 2 , 3 , 2 , 3 ], [ 5 , 5 , 4 , 7 , 7 , 4 ], [ 1 , 2 ], [ 1 , 1 , 2 , 2 , 2 ]]
print ( "The original list is : " + str (test_list))
K = 2
test_list.sort(key = get_Kfreq)
print ( "Sorted List : " + str (test_list))
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Output:
The original list is : [[10, 2, 3, 2, 3], [5, 5, 4, 7, 7, 4], [1, 2], [1, 1, 2, 2, 2]] Sorted List : [[5, 5, 4, 7, 7, 4], [1, 2], [10, 2, 3, 2, 3], [1, 1, 2, 2, 2]]
Time Complexity: O(m*nlogn)
Auxiliary Space: O(k)
Method #2 : Using sorted() + lambda + count()
In this, we perform the task of sorting using sorted() and lambda, eliminates the external function call and lambda function used for computation.
Python3
test_list = [[ 10 , 2 , 3 , 2 , 3 ], [ 5 , 5 , 4 , 7 , 7 , 4 ], [ 1 , 2 ], [ 1 , 1 , 2 , 2 , 2 ]]
print ( "The original list is : " + str (test_list))
K = 2
res = sorted (test_list, key = lambda row: row.count(K))
print ( "Sorted List : " + str (res))
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Output:
The original list is : [[10, 2, 3, 2, 3], [5, 5, 4, 7, 7, 4], [1, 2], [1, 1, 2, 2, 2]] Sorted List : [[5, 5, 4, 7, 7, 4], [1, 2], [10, 2, 3, 2, 3], [1, 1, 2, 2, 2]]
Time Complexity: O(nlogn), where n is the number of elements in the list “test_list”.
Auxiliary Space: O(n), where n is the number of elements in the list “test_list”.
Method #3 : Using operator.countOf()
Python3
import operator as op
def get_Kfreq(row):
return op.countOf(row,K)
test_list = [[ 10 , 2 , 3 , 2 , 3 ], [ 5 , 5 , 4 , 7 , 7 , 4 ], [ 1 , 2 ], [ 1 , 1 , 2 , 2 , 2 ]]
print ( "The original list is : " + str (test_list))
K = 2
test_list.sort(key = get_Kfreq)
print ( "Sorted List : " + str (test_list))
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Output
The original list is : [[10, 2, 3, 2, 3], [5, 5, 4, 7, 7, 4], [1, 2], [1, 1, 2, 2, 2]]
Sorted List : [[5, 5, 4, 7, 7, 4], [1, 2], [10, 2, 3, 2, 3], [1, 1, 2, 2, 2]]
Time Complexity: O(N*M)
Auxiliary Space: O(N*M)
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