Python – Filter Tuples by Kth element from List

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Given a list of tuples, filter by Kth element presence in List.

Input : test_list = [(“GFg”, 5, 9), (“is”, 4, 3), (“best”, 10, 29)], check_list = [4, 2, 3, 10], K = 2
Output : [(‘is’, 4, 3)]
Explanation : 3 is 2nd element and present in list, hence filtered tuple.

Input : test_list = [(“GFg”, 5, 9), (“is”, 4, 3), (“best”, 10, 29)], check_list = [4, 2, 3, 10], K = 1
Output : [(‘is’, 4, 3), (‘best’, 10, 29)]
Explanation : 4 and 10 are 1st elements and present in list, hence filtered tuples.

Method #1: Using list comprehension

In this, we check for each element of Tuple’s Kth element to be present in list in shorthand using list comprehension and containment is tested using in operator.

Python3

# Python3 code to demonstrate working of 
# Filter Tuples by Kth element from List
# Using list comprehension
 
# initializing list
test_list = [("GFg", 5, 9), ("is", 4, 3), ("best", 10, 29)]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing check_list
check_list = [4, 2, 8, 10]
 
# initializing K 
K = 1
 
# checking for presence on Kth element in list 
# one liner 
res = [sub for sub in test_list if sub[K] in check_list]
 
# printing result 
print("The filtered tuples : " + str(res))

Output

The original list is : [('GFg', 5, 9), ('is', 4, 3), ('best', 10, 29)]
The filtered tuples : [('is', 4, 3), ('best', 10, 29)]

Time complexity: O(n), where n is the length of the test_list.
Auxiliary space: O(m), where m is the length of the resulting list after filtering.

Method #2 : Using filter() + lambda

In this, lambda function checks for element presence and filter performs task of filtering tuples.

Python3

# Python3 code to demonstrate working of 
# Filter Tuples by Kth element from List
# Using filter() + lambda
 
# initializing list
test_list = [("GFg", 5, 9), ("is", 4, 3), ("best", 10, 29)]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing check_list
check_list = [4, 2, 8, 10]
 
# initializing K 
K = 1
 
# filter() perform filter, lambda func. checks for presence
# one liner 
res = list(filter(lambda sub: sub[K] in check_list, test_list))
 
# printing result 
print("The filtered tuples : " + str(res))

Output

The original list is : [('GFg', 5, 9), ('is', 4, 3), ('best', 10, 29)]
The filtered tuples : [('is', 4, 3), ('best', 10, 29)]

Time complexity: O(n), where n is the length of the test_list
Auxiliary space: O(m), where m is the length of the check_list.

Method #3: Using for loop

Steps:

Initialize the test_list and check_list variables.
Initialize K to the index of the Kth element in the tuples.
Initialize an empty list called res to hold the filtered tuples.
Iterate over the tuples in test_list using a for loop.
Check if the Kth element of the current tuple is in check_list.
If it is, append the tuple to res.
Print the filtered tuples.

Python3

# Python3 code to demonstrate working of 
# Filter Tuples by Kth element from List
# Using for loop
 
# initializing list
test_list = [("GFg", 5, 9), ("is", 4, 3), ("best", 10, 29)]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing check_list
check_list = [4, 2, 8, 10]
 
# initializing K 
K = 1
 
# initializing empty result list
res = []
 
# iterating over tuples in test_list
for tup in test_list:
    # checking if Kth element is in check_list
    if tup[K] in check_list:
        # appending tuple to result list
        res.append(tup)
 
# printing result 
print("The filtered tuples : " + str(res))

Output

The original list is : [('GFg', 5, 9), ('is', 4, 3), ('best', 10, 29)]
The filtered tuples : [('is', 4, 3), ('best', 10, 29)]

Time Complexity: O(n), where n is the number of tuples in test_list.

Auxiliary Space: O(k), where k is the number of tuples that pass the filter. The space used by the res list.

Method#4: Using the Recursive method:

Algorithm:

Check if the input list `test_list` is empty or not, If yes, then return an empty list.
If not, check if the Kth element of the first tuple in `test_list` is in the `check_list`.
If it is, append the first tuple to the result list and recursively call the function on the remaining part of the list (`test_list[1:]`) with the same `K` and `check_list` arguments.
If it is not, recursively call the function on the remaining part of the list (`test_list[1:]`) with the same `K` and `check_list` arguments.
Return the result list obtained from steps 3 and 4.

Below is the implementation of the above approach:

Python3

# Python program for the above approach
 
# Function to filter tuples
def filter_tuples(test_list, K, check_list):
    if not test_list:
        return []
    if test_list[0][K] in check_list:
        return [test_list[0]] + filter_tuples(test_list[1:], K, check_list)
    else:
        return filter_tuples(test_list[1:], K, check_list)
 
 
# initializing list
test_list = [("GFg", 5, 9), ("is", 4, 3), ("best", 10, 29)]
 
# initializing check_list
check_list = [4, 2, 8, 10]
 
# initializing K
K = 1
 
# calling function and storing result in res
res = filter_tuples(test_list, K, check_list)
 
# printing original list
print("The original list is : " + str(test_list))
 
# printing result
print("The filtered tuples : " + str(res))

Output

The original list is : [('GFg', 5, 9), ('is', 4, 3), ('best', 10, 29)]
The filtered tuples : [('is', 4, 3), ('best', 10, 29)]

The time complexity of this recursive function is O(n), where n is the number of tuples in the input list. This is because the function visits each tuple in the list once.

The space complexity is also O(n) because the function creates a new list to store the filtered tuples. However, in the worst case where all tuples in the input list are selected, the space complexity can be O(n^2) due to the recursive calls creating new lists.