Python Program to create a sub-dictionary containing all keys from dictionary list

Given the dictionary list, our task is to create a new dictionary list that contains all the keys, if not, then assign None to the key and persist of each dictionary.

Example:

Input : test_list = [{‘gfg’ : 3, ‘is’ : 7}, {‘gfg’ : 3, ‘is’ : 1, ‘best’ : 5}, {‘gfg’ : 8}]
Output : [{‘is’: 7, ‘best’: None, ‘gfg’: 3}, {‘is’: 1, ‘best’: 5, ‘gfg’: 3}, {‘is’: None, ‘best’: None, ‘gfg’: 8}]
Explanation : The items with “is” and “best” are added to all lists, wherever missing as None if no values populated.

Input : test_list = [{‘gfg’ : 3}, {‘gfg’ : 3, ‘best’ : 5}, {‘gfg’ : 8}]
Output : [{‘best’: None, ‘gfg’: 3}, {‘best’: 5, ‘gfg’: 3}, {‘best’: None, ‘gfg’: 8}]
Explanation : The items with “best” are added to all lists, wherever missing as None if no values populated.

Method #1 : Using set() + chain.from_iterable() + get() + list comprehension

In this, we perform the task of getting all the required keys using set() and chain.from_iterable(). The next step is to update all the dictionaries with not found keys using list comprehension and get().

Output:

The original list is : [{‘gfg’: 3, ‘is’: 7}, {‘gfg’: 3, ‘is’: 1, ‘best’: 5}, {‘gfg’: 8}]

Reformed dictionaries list : [{‘gfg’: 3, ‘best’: None, ‘is’: 7}, {‘gfg’: 3, ‘best’: 5, ‘is’: 1}, {‘gfg’: 8, ‘best’: None, ‘is’: None}]

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

Method #2 : Using set() + chain.from_iterable() + update()

In this, the updation and checking of all the keys from dictionary is done using update(), rest all the functions remain similar. 

Step-by-step approach:

• Import the required modules: chain from the itertools module.
• Initialize a list of dictionaries named test_list with some sample data.
• Print the original list using the print() function.
• Use the chain.from_iterable() function to flatten the list of dictionaries into a single iterable.
• Convert the iterable into a set using the set() function, to obtain all the unique keys in the list of dictionaries, and store them in the variable all_keys.
• Iterate over each dictionary in test_list using a for loop.
  • Use the update() method to add a new key-value pair to each dictionary for each key in all_keys that is not already in the current dictionary. The new value assigned is None.
• Print the updated list of dictionaries using the print() function.

Below is the implementation of the above approach:

Output:

The original list is : [{‘gfg’: 3, ‘is’: 7}, {‘gfg’: 3, ‘is’: 1, ‘best’: 5}, {‘gfg’: 8}]

Reformed dictionaries list : [{‘gfg’: 3, ‘best’: None, ‘is’: 7}, {‘gfg’: 3, ‘best’: 5, ‘is’: 1}, {‘gfg’: 8, ‘best’: None, ‘is’: None}]

Time complexity: O(n*m).
Auxiliary space: O(n*m).

Method 3 : set() + dictionary comprehension

Step-by-step approach:

• First, initialize the test_list with some dictionaries.
• Extract all the keys from the dictionaries using set() and for loop in a generator expression.
• Create a new list of dictionaries using dictionary comprehension. For each dictionary in test_list, we iterate over all the keys in all_keys and use the get() method to retrieve the value for that key. If the key is missing, we assign None to it.
• Finally, print the resulting list of dictionaries.

Below is the implementation of the above approach:

Time complexity: O(nm), where n is the length of test_list and m is the average number of keys per dictionary.
Auxiliary space: O(nm), since we create a new list of dictionaries with the same number of elements as test_list, and each dictionary can have up to m keys.

Method 4: Using zip() and dict() constructor

1. Use the zip() function to transpose test_list into a list of tuples, where each tuple contains the values for each key.
2. Use the set() function to get all unique keys from test_list.
3. Initialize a list of dictionaries test_list_dict with the same number of dictionaries as in test_list, but with all values set to None.
4. Use a for loop to iterate through the tuples of values and keys simultaneously.
5. Use the dict() constructor to create a new dictionary with the values and keys from each tuple, and update the corresponding dictionary in test_list_dict.
6. Return test_list_dict.

Time complexity: O(NK), where n is the number of dictionaries in test_list and k is the average number of keys per dictionary.
Auxiliary space: O(k), where k is the total number of unique keys in test_list.

Method 5: Using defaultdict(): The idea is to use defaultdict() from the collections module to create a new dictionary with the default value None and then update it with the key-value pairs from the original dictionaries. Below are the steps:

1. Import the collections module.
2. Initialize a defaultdict object with default value None.
3. Loop over each dictionary in the list and update the defaultdict with its key-value pairs using the update() method.
4. Convert the defaultdict to a regular dictionary using the dict() constructor.
5. Append the resulting dictionary to a new list.
6. Print the final list.

Below is the implementation of the above approach:

Time Complexity: O(N*M), where N is the number of dictionaries in the list and M is the maximum number of keys in any dictionary.
Auxiliary space: O(M), where M is the maximum number of keys in any dictionary.