Python – Remove nested records from tuple

Last Updated : 27 Apr, 2023

Sometimes, while working with records, we can have a problem in which an element of a record is another tuple records and we might have to remove the nested records. This is a problem which does not occur commonly, but having a solution to it is useful. Let’s discuss certain way in which this task can be performed.

Method 1: Using loop + isinstance() + enumerate()
This problem can be solved using the above functionalities. In this, we just loop through the elements using enumerate() to get the index count of it and check the type using isinstance() and recreate the new tuple by checking ignoring tuple records.

Python3

# Python3 code to demonstrate working of
# Remove nested records
# using isinstance() + enumerate() + loop
 
# initialize tuple
test_tup = (1, 5, 7, (4, 6), 10)
 
# printing original tuple
print("The original tuple : " + str(test_tup))
 
# Remove nested records
# using isinstance() + enumerate() + loop
res = tuple()
for count, ele in enumerate(test_tup):
    if not isinstance(ele, tuple):
        res = res + (ele, )
 
# printing result
print("Elements after removal of nested records : " + str(res))

Output

The original tuple : (1, 5, 7, (4, 6), 10)
Elements after removal of nested records : (1, 5, 7, 10)

Time Complexity: O(n), where n is the length of the list test_tup
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the res list

Method 2 : Using type() method

Python3

# Python3 code to demonstrate working of
# Remove nested records
 
# initialize tuple
test_tup = (1, 5, 7, (4, 6), 10)
 
# printing original tuple
print("The original tuple : " + str(test_tup))
 
# Remove nested records
res=[]
for i in test_tup:
    if not type(i) is tuple:
        res.append(i)
res=tuple(res)
# printing result
print("Elements after removal of nested records : " + str(res))

Output

The original tuple : (1, 5, 7, (4, 6), 10)
Elements after removal of nested records : (1, 5, 7, 10)

Time complexity: O(n), where n is the number of elements in the tuple.

Auxiliary Space: O(n), where n is the number of elements in the tuple.

Method 3 : Using filter()+lambda functions

Python3

# Python3 code to demonstrate working of
# Remove nested records
 
# initialize tuple
test_tup = (1, 5, 7, (4, 6), 10)
 
# printing original tuple
print("The original tuple : " + str(test_tup))
 
# Remove nested records
res = list(filter(lambda x: not isinstance(x, tuple), test_tup))
 
# printing result
print("Elements after removal of nested records : " + str(res))

Output

The original tuple : (1, 5, 7, (4, 6), 10)
Elements after removal of nested records : [1, 5, 7, 10]

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

Method 4: Using list comprehension

Python3

# Python3 code to demonstrate working of
# Remove nested records
  
# initialize tuple
test_tup = (1, 5, 7, (4, 6), 10)
  
# printing original tuple
print("The original tuple : " + str(test_tup))
  
# Remove nested records
res = [x for x in test_tup if not isinstance(x, tuple)]
  
# printing result
print("Elements after removal of nested records : " + str(res))
#This code is contributed by Edula Vinay Kumar Reddy

Output

The original tuple : (1, 5, 7, (4, 6), 10)
Elements after removal of nested records : [1, 5, 7, 10]

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

Python3

from functools import reduce
test_tup = (1, 5, 7, (4, 6), 10)
# printing original tuple
print("The original tuple : " + str(test_tup))
res = reduce(lambda acc, x: acc + (x,) if not isinstance(x, tuple) else acc, test_tup, ())
print(res)
#This code is contributed by Jyothi pinjala.

Output

The original tuple : (1, 5, 7, (4, 6), 10)
(1, 5, 7, 10)

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

Method 5: Method 5: Using itertools.chain()

Use the itertools.chain() function to flatten the nested tuple and create a new tuple without the inner tuples.

Python3

import itertools
 
test_tup = (1, 5, 7, (4, 6), 10)
print("The original tuple: " + str(test_tup))
 
res = tuple(itertools.chain(*([x] if not isinstance(x, tuple) else x for x in test_tup)))
print(res)

Output

The original tuple: (1, 5, 7, (4, 6), 10)
(1, 5, 7, 4, 6, 10)

Time complexity: O(n), where n is the number of elements in the tuple.
Auxiliary space: O(n), as it creates a new tuple of the same length as the original tuple.

Method 6: using recursion

Define a function flatten_tuple that takes a tuple as an argument.
Create an empty list result.
For each element elem in the input tuple, check if it is a tuple using isinstance(elem, tuple).
If elem is not a tuple, append it to the result list.
If elem is a tuple, recursively call the flatten_tuple function on it and extend the result list with the flattened tuple.
Return the flattened result list as a tuple.

Python3

def flatten_tuple(tup):
    """
    Recursively flatten a tuple of any depth into a single tuple.
 
    Args:
        tup: A tuple to be flattened.
 
    Returns:
        A flattened tuple.
 
    """
    result = []
    for elem in tup:
        if not isinstance(elem, tuple):
            result.append(elem)
        else:
            result.extend(flatten_tuple(elem))
    return tuple(result)
 
# Driver code to test the function
test_tup = (1, 5, 7, (4, 6), 10)
print("The original tuple: " + str(test_tup))
 
# Call the function to flatten the tuple
res = flatten_tuple(test_tup)
 
# Print the flattened tuple
print(res)

Output

The original tuple: (1, 5, 7, (4, 6), 10)
(1, 5, 7, 4, 6, 10)

Time complexity: O(n), where n is the length of given test_tup
Auxiliary space: O(n) since it creates a new list to store the flattened elements.