Given a dictionary, replace None values in every nesting with an empty dictionary.
Input : test_dict = {“Gfg” : {1 : None, 7 : None}, “is” : None, “Best” : [1, { 5 : None }, 9, 3]}
Output : {‘Gfg’: {1: {}, 7: {}}, ‘is’: {}, ‘Best’: [1, {5: {}}, 9, 3]}
Explanation : All None values are replaced by empty dictionaries.Input : test_dict = {“Gfg” : {7 : None}, “is” : None, “Best” : [1, { 5 : None }, 9, 3]}
Output : {‘Gfg’: {7: {}}, ‘is’: {}, ‘Best’: [1, {5: {}}, 9, 3]}
Explanation : All None values are replaced by empty dictionaries.
Method : Using recursion + isinstance()
In this, we check for dictionary instance using isinstance() and call for recursion for nested dictionary replacements. This also checks for nested instances in form of list elements and checks for the list using isinstance().
Python3
# Python3 code to demonstrate working of# Replace None with Empty Dictionary# Using recursion + isinstance()# helper function to perform taskdef replace_none(test_dict): # checking for dictionary and replacing if None if isinstance(test_dict, dict): for key in test_dict: if test_dict[key] is None: test_dict[key] = {} else: replace_none(test_dict[key]) # checking for list, and testing for each value elif isinstance(test_dict, list): for val in test_dict: replace_none(val)# initializing dictionarytest_dict = {"Gfg": {1: None, 7: 4}, "is": None, "Best": [1, {5: None}, 9, 3]}# printing original dictionaryprint("The original dictionary is : " + str(test_dict))# calling helper fncreplace_none(test_dict)# printing resultprint("The converted dictionary : " + str(test_dict)) |
Output:
The original dictionary is : {‘Gfg’: {1: None, 7: 4}, ‘is’: None, ‘Best’: [1, {5: None}, 9, 3]} The converted dictionary : {‘Gfg’: {1: {}, 7: 4}, ‘is’: {}, ‘Best’: [1, {5: {}}, 9, 3]}
Method 2: Using stack
Step-by-step approach:
- Start by creating an empty stack and push the input dictionary onto it.
- While the stack is not empty, pop the top item from the stack.
- If the popped item is a dictionary, loop through its key-value pairs:
a. If the value is None, replace it with an empty dictionary.
b. If the value is a dictionary or a list, push it onto the stack.
If the popped item is a list, loop through its elements:
a. If the element is None, replace it with an empty dictionary.
b. If the element is a dictionary or a list, push it onto the stack. - Repeat steps 2-4 until the stack is empty.
- Return the modified dictionary.
Below is the implementation of the above approach:
Python3
def replace_none(test_dict): # Create a stack with the initial dictionary stack = [test_dict] # While the stack is not empty, process the dictionaries in the stack while stack: cur_dict = stack.pop() # If the current item in the stack is a dictionary, process its key-value pairs if isinstance(cur_dict, dict): for key, val in cur_dict.items(): # If the value is None, replace it with an empty dictionary if val is None: cur_dict[key] = {} # If the value is a dictionary or a list, push it onto the stack elif isinstance(val, (dict, list)): stack.append(val) # If the current item in the stack is a list, process its elements elif isinstance(cur_dict, list): for i, val in enumerate(cur_dict): # If the value is None, replace it with an empty dictionary if val is None: cur_dict[i] = {} # If the value is a dictionary or a list, push it onto the stack elif isinstance(val, (dict, list)): stack.append(val)# Example usagetest_dict = {"Gfg": {1: None, 7: 4}, "is": None, "Best": [1, {5: None}, 9, 3]}replace_none(test_dict)print(test_dict) |
{'Gfg': {1: {}, 7: 4}, 'is': {}, 'Best': [1, {5: {}}, 9, 3]}
Time complexity: O(n)
Auxiliary space: O(n)
