Python – Remove Tuples of Length K

Last Updated : 02 May, 2023

Given list of tuples, remove all the tuples with length K.

Input : test_list = [(4, 5), (4, ), (8, 6, 7), (1, ), (3, 4, 6, 7)], K = 2
Output : [(4, ), (8, 6, 7), (1, ), (3, 4, 6, 7)]
Explanation : (4, 5) of len = 2 is removed.

Input : test_list = [(4, 5), (4, ), (8, 6, 7), (1, ), (3, 4, 6, 7)], K = 3
Output : [(4, 5), (4, ), (1, ), (3, 4, 6, 7)]
Explanation : 3 length tuple is removed.

Method #1 : Using list comprehension

This is one of the ways in which this task can be performed. In this, we iterate for all elements in loop and perform the required task of removal of K length elements using conditions.

Python3

# Python3 code to demonstrate working of 
# Remove Tuples of Length K
# Using list comprehension
 
# initializing list
test_list = [(4, 5), (4, ), (8, 6, 7), (1, ), (3, 4, 6, 7)]
 
# printing original list
print("The original list : " + str(test_list))
 
# initializing K 
K = 1
 
# 1 liner to perform task
# filter just lengths other than K 
# len() used to compute length
res = [ele for ele in test_list if len(ele) != K]
 
# printing result 
print("Filtered list : " + str(res))

Output

The original list : [(4, 5), (4, ), (8, 6, 7), (1, ), (3, 4, 6, 7)]
Filtered list : [(4, 5), (8, 6, 7), (3, 4, 6, 7)]

Time Complexity: O(n), where n is the number of tuples in the input list.
Auxiliary Space: O(m), where m is the number of tuples in the output list (since we are creating a new list to store the filtered tuples).

Method #2 : Using filter() + lambda + len()

Yet another way to solve this problem. In this, we perform filtering using filter() and lambda function to extract just non K length elements using len().

Python3

# Python3 code to demonstrate working of 
# Remove Tuples of Length K
# Using filter() + lambda + len() 
 
# initializing list
test_list = [(4, 5), (4, ), (8, 6, 7), (1, ), (3, 4, 6, 7)]
 
# printing original list
print("The original list : " + str(test_list))
 
# initializing K 
K = 1
 
# filter() filters non K length values and returns result
res = list(filter(lambda x : len(x) != K, test_list))
 
# printing result 
print("Filtered list : " + str(res))

Output

The original list : [(4, 5), (4, ), (8, 6, 7), (1, ), (3, 4, 6, 7)]
Filtered list : [(4, 5), (8, 6, 7), (3, 4, 6, 7)]

Time complexity: O(n), where n is the length of the input list.
Auxiliary space: O(n), as a new list is created to store the filtered values.

Method #3: Using a loop

Use a for loop to iterate through the list and remove tuples of length K.

Python3

# Python3 code to demonstrate working of 
# Remove Tuples of Length K
# Using list comprehension
 
# initializing list
test_list = [(4, 5), (4, ), (8, 6, 7), (1, ), (3, 4, 6, 7)]
 
# printing original list
print("The original list : " + str(test_list))
 
# initializing K 
K = 1
 
# using list comprehension to filter out tuples of length K
# len() is used to compute length
res = [ele for ele in test_list if len(ele) != K]
 
# printing result 
print("Filtered list : " + str(res))

Output

The original list : [(4, 5), (4,), (8, 6, 7), (1,), (3, 4, 6, 7)]
Filtered list : [(4, 5), (8, 6, 7), (3, 4, 6, 7)]

Time complexity: O(n), where n is the length of the input list.
Auxiliary space: O(n)

METHOD 4:Using map() and lambda function

APPROACH:

The given problem is to remove tuples of a given length k from a list of tuples and return the filtered list.

ALGORITHM:

1.Read the original list and the value of k.
2.Create an empty list to store the filtered tuples.
3.Use the map() function with a lambda function as its first argument to iterate over each tuple in the original list.
4.Use the filter() function with a lambda function as its first argument to filter out the tuples whose length is equal to k.
5.Convert the resulting map object into a list using the list() function and store it in the filtered_list.
6.Print the filtered_list.

Python3

# input
original_list = [(4, 5), (4, ), (8, 6, 7), (1, ), (3, 4, 6, 7)]
k = 1
 
# use map() and a lambda function to filter out tuples of length k
filtered_list = list(map(lambda x: x, filter(lambda x: len(x) != k, original_list)))
 
# output
print(filtered_list)

Output

[(4, 5), (8, 6, 7), (3, 4, 6, 7)]

Time Complexity:
The time complexity of this approach is O(n), where n is the number of tuples in the original list. This is because we are iterating over each tuple in the original list only once.

Space Complexity:
The space complexity of this approach is also O(n), where n is the number of tuples in the original list. This is because we are creating a new list to store the filtered tuples.

METHOD 5 :Using heapq:

Algorithm:

Initialize an empty result list.
Traverse the input list and for each tuple:
a. Check if the length of the tuple is not equal to K.
b. If the length is not equal to K, add the tuple to the result list.
Return the result list.

Python3

import heapq
 
# initializing list
test_list = [(4, 5), (4, ), (8, 6, 7), (1, ), (3, 4, 6, 7)]
# printing original list
print("The original list : " + str(test_list))
# initializing K
K = 1
 
# filtering non K length values using heapq
res = list(filter(lambda x: len(x) != K, test_list))
 
# printing result
print("Filtered list : " + str(res))
#This code is contributed by Vinay Pinjala.

Output

The original list : [(4, 5), (4,), (8, 6, 7), (1,), (3, 4, 6, 7)]
Filtered list : [(4, 5), (8, 6, 7), (3, 4, 6, 7)]

Time Complexity: O(N), where N is the number of tuples in the input list. We traverse the list once to filter the tuples.

Auxiliary Space: O(N), where N is the number of tuples in the input list. We create a new list to store the filtered tuples. The size of this list is equal to or less than the size of the input list, depending on the number of tuples that satisfy the condition.