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Python – Remove Tuples from the List having every element as None

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Given a Tuple list, remove all tuples with all None values.

Input : test_list = [(None, 2), (None, None), (3, 4), (12, 3), (None, )] 
Output : [(None, 2), (3, 4), (12, 3)] 
Explanation : All None tuples are removed.
Input : test_list = [(None, None), (None, None), (3, 4), (12, 3), (None, )] 
Output : [(3, 4), (12, 3)] 
Explanation : All None tuples are removed. 

Method #1 : Using all() + list comprehension

In this, we use all() to check for all None values for discarding and list comprehension does task of iteration.

Python3

# Python3 code to demonstrate working of
# Remove None Tuples from List
# Using all() + list comprehension
 
# initializing list
test_list = [(None, 2), (None, None), (3, 4), (12, 3), (None, )]
 
# printing original list
print("The original list is : " + str(test_list))
 
# negating result for discarding all None Tuples
res = [sub for sub in test_list if not all(ele == None for ele in sub)]
 
# printing result
print("Removed None Tuples : " + str(res))

                    

Output
The original list is : [(None, 2), (None, None), (3, 4), (12, 3), (None,)]
Removed None Tuples : [(None, 2), (3, 4), (12, 3)]

Time complexity: O(n), where n is the length of the input list.
Auxiliary space: O(m), where m is the number of non-None tuples in the input list.

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

In this method, task of filtering None tuples is done using filter() and lambda function to provide None checking functionality using all().

Python3

# Python3 code to demonstrate working of
# Remove None Tuples from List
# Using filter() + lambda + all()
 
# initializing list
test_list = [(None, 2), (None, None), (3, 4), (12, 3), (None, )]
 
# printing original list
print("The original list is : " + str(test_list))
 
# filter() + lambda to drive logic of discarding tuples
res = list(filter(lambda sub : not all(ele == None for ele in sub), test_list))
 
# printing result
print("Removed None Tuples : " + str(res))

                    

Output
The original list is : [(None, 2), (None, None), (3, 4), (12, 3), (None,)]
Removed None Tuples : [(None, 2), (3, 4), (12, 3)]

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

Method #3: Using count() method

If count of None in each tuple equals to length of tuple, then the entire tuple elements are None.Used this in if condition and appended other tuples to output list

Python3

# Python3 code to demonstrate working of
# Remove None Tuples from List
 
# initializing list
test_list = [(None, None), (None, None), (3, 4), (12, 3), (None, )]
 
# printing original list
print("The original list is : " + str(test_list))
 
# negating result for discarding all None Tuples
res=[]
for i in test_list:
    if not(i.count(None)== len(i)):
        res.append(i)
 
# printing result
print("Removed None Tuples : " + str(res))

                    

Output
The original list is : [(None, None), (None, None), (3, 4), (12, 3), (None,)]
Removed None Tuples : [(3, 4), (12, 3)]

Time Complexity: O(n)
Auxiliary Space: O(n)

Method #4: Using enumerate function

Python3

test_list = [(None, None), (None, None), (3, 4), (12, 3), (None, )] 
res = [sub for i,sub in enumerate(test_list) if not all(ele == None for ele in sub)]
print(res)

                    

Output
[(3, 4), (12, 3)]

Time Complexity: O(n) where n is the number of elements in the list “test_list”.  enumerate function performs n number of operations.
Auxiliary Space: O(n), extra space is required 

Method #5 : Using operator.countOf()

Python3

# Python3 code to demonstrate working of
# Remove None Tuples from List
 
# initializing list
test_list = [(None, None), (None, None), (3, 4), (12, 3), (None, )]
 
# printing original list
print("The original list is : " + str(test_list))
 
# negating result for discarding all None Tuples
res=[]
import operator
for i in test_list:
    if not(operator.countOf(i,None)== len(i)):
        res.append(i)
 
# printing result
print("Removed None Tuples : " + str(res))

                    

Output
The original list is : [(None, None), (None, None), (3, 4), (12, 3), (None,)]
Removed None Tuples : [(3, 4), (12, 3)]

Time Complexity : O(N)
Auxiliary Space : O(N)

Method 6: Using a for loop and slicing

In this method, we use a for loop to iterate through the list and remove the None tuples by slicing the list. 

Python3

# Python3 code to demonstrate working of
# Remove None Tuples from List
 
# initializing list
test_list = [(None, None), (None, None), (3, 4), (12, 3), (None, )]
 
# printing original list
print("The original list is : " + str(test_list))
 
# using for loop and slicing to remove None Tuples
res = []
for i in range(len(test_list)):
    if None not in test_list[i]:
        res.append(test_list[i])
 
# printing result
print("Removed None Tuples : " + str(res))

                    

Output
The original list is : [(None, None), (None, None), (3, 4), (12, 3), (None,)]
Removed None Tuples : [(3, 4), (12, 3)]

Time Complexity: O(n), where n is the length of the list
Auxiliary Space: O(n), where n is the length of the list (to store the result list)

Method 7: Using recursion

Step-by-step approach:

  • Define the original list
  • Define a recursive function that takes the original list and an empty list as arguments
  • The function should check if the original list is empty
  • If it is, return the empty list
  • Otherwise, check if the first element of the list is a tuple containing None values
  • If it is, call the function recursively with the remaining elements of the list and the empty list
  • Otherwise, append the first element of the list to the empty list and call the function recursively with the remaining elements of the list and the updated list
  • Call the recursive function with the original list and an empty list as arguments and assign the result to a new variabl

Python3

# Define the original list
test_list = [(None, None), (None, None), (3, 4), (12, 3), (None, )]
 
# Define the recursive function
def remove_none_tuples(lst, res):
    if not lst:
        return res
    elif None in lst[0]:
        return remove_none_tuples(lst[1:], res)
    else:
        res.append(lst[0])
        return remove_none_tuples(lst[1:], res)
 
# Call the recursive function with the original list and an empty list
new_list = remove_none_tuples(test_list, [])
 
# Print the new list
print("Removed None Tuples : " + str(new_list))

                    

Output
Removed None Tuples : [(3, 4), (12, 3)]

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



Last Updated : 15 May, 2023
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