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 tupletest_tup = (1, 5, 7, (4, 6), 10)# printing original tupleprint("The original tuple : " + str(test_tup))# Remove nested records# using isinstance() + enumerate() + loopres = tuple()for count, ele in enumerate(test_tup): if not isinstance(ele, tuple): res = res + (ele, )# printing resultprint("Elements after removal of nested records : " + str(res)) |
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 tupletest_tup = (1, 5, 7, (4, 6), 10)# printing original tupleprint("The original tuple : " + str(test_tup))# Remove nested recordsres=[]for i in test_tup: if not type(i) is tuple: res.append(i)res=tuple(res)# printing resultprint("Elements after removal of nested records : " + str(res)) |
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 tupletest_tup = (1, 5, 7, (4, 6), 10)# printing original tupleprint("The original tuple : " + str(test_tup))# Remove nested recordsres = list(filter(lambda x: not isinstance(x, tuple), test_tup))# printing resultprint("Elements after removal of nested records : " + str(res)) |
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 tupletest_tup = (1, 5, 7, (4, 6), 10) # printing original tupleprint("The original tuple : " + str(test_tup)) # Remove nested recordsres = [x for x in test_tup if not isinstance(x, tuple)] # printing resultprint("Elements after removal of nested records : " + str(res))#This code is contributed by Edula Vinay Kumar Reddy |
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 reducetest_tup = (1, 5, 7, (4, 6), 10)# printing original tupleprint("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. |
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 itertoolstest_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) |
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 functiontest_tup = (1, 5, 7, (4, 6), 10)print("The original tuple: " + str(test_tup))# Call the function to flatten the tupleres = flatten_tuple(test_tup)# Print the flattened tupleprint(res) |
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.
