Python program to update a dictionary with the values from a dictionary list

Given a dictionary and dictionary list, update the dictionary with dictionary list values.

Input : test_dict = {“Gfg” : 2, “is” : 1, “Best” : 3}, dict_list = [{‘for’ : 3, ‘all’ : 7}, {‘and’ : 1, ‘CS’ : 9}] 
Output : {‘Gfg’: 2, ‘is’: 1, ‘Best’: 3, ‘for’: 3, ‘all’: 7, ‘and’: 1, ‘CS’: 9} 
Explanation : All dictionary keys updated in single dictionary.

Input : test_dict = {“Gfg” : 2, “is” : 1, “Best” : 3}, dict_list = [{‘for’ : 3, ‘all’ : 7}] 
Output : {‘Gfg’: 2, ‘is’: 1, ‘Best’: 3, ‘for’: 3, ‘all’: 7} 
Explanation : All dictionary keys updated in single dictionary. 

Method #1 : Using update() + loop

In this, we iterate through all the elements in loop, and perform update to update all the keys in dictionary to original dictionary.

Step-by-step approach :

• First, initialize a dictionary test_dict with some key-value pairs.
• Print the original dictionary.
• Initialize a list of dictionaries dict_list.
• Use a for loop to iterate through the dict_list.
• Inside the loop, use the update() method to update the test_dict with the current dictionary dicts in the loop.
• Print the updated dictionary test_dict.

Below is the implementation of the above approach:

The original dictionary is : {'Gfg': 2, 'is': 1, 'Best': 3}
The updated dictionary : {'Gfg': 2, 'is': 1, 'Best': 3, 'for': 3, 'all': 7, 'Lazyroar': 10, 'and': 1, 'CS': 9}

Time Complexity: O(n)
Auxiliary Space: O(1)

Method #2 : Using ChainMap + ** operator

In this, we perform task of merging all list dictionaries into 1 using ChainMap and ** operator is used to merge target dictionary to merged dictionary.

The original dictionary is : {'Gfg': 2, 'is': 1, 'Best': 3}
The updated dictionary : {'Gfg': 2, 'is': 1, 'Best': 3, 'and': 1, 'CS': 9, 'Lazyroar': 10, 'for': 3, 'all': 7}

Time Complexity: O(n)
Auxiliary Space: O(1)

Method 3: Using dictionary comprehension.

Using dictionary comprehension to create a new dictionary from the list of dictionaries. We are using the unpacking operator (**), which unpacks the key-value pairs from both dictionaries into a single dictionary.

Follow the below steps to implement the above idea:

• Initialize the original dictionary.
• Print the original dictionary.
• Initialize the list of dictionaries.
• Use dictionary comprehension to create a new dictionary by iterating through each dictionary in the list of dictionaries and unpacking its key-value pairs.
• Use the unpacking operator (**) to merge the new dictionary with the original dictionary and create the updated dictionary.
• Print the updated dictionary.

Below is the implementation of the above approach:

The original dictionary is : {'Gfg': 2, 'is': 1, 'Best': 3}
The updated dictionary : {'Gfg': 2, 'is': 1, 'Best': 3, 'for': 3, 'all': 7, 'Lazyroar': 10, 'and': 1, 'CS': 9}

Time complexity: O(nm), where n is the length of the list of dictionaries and m is the average number of key-value pairs in each dictionary.
Auxiliary space: O(nm), as we are creating a new dictionary to store the updated key-value pairs.

Method 4: Using a list comprehension and the update() 

Step-by-step approach:

• Create the original test_dict dictionary with 3 key-value pairs.
• Print the original dictionary using the print() function and the str() method to convert the dictionary to a string.
• Create a list of dictionaries named dict_list with 3 dictionaries containing key-value pairs.
• Use a list comprehension to iterate over each dictionary in the dict_list and update the test_dict with the update() method.
• Print the updated dictionary using the print() function and the str() method to convert the dictionary to a string.

The original dictionary is: {'Gfg': 2, 'is': 1, 'Best': 3}
The updated dictionary: {'Gfg': 2, 'is': 1, 'Best': 3, 'for': 3, 'all': 7, 'Lazyroar': 10, 'and': 1, 'CS': 9}

Time complexity: O(n*m), where n is the length of the dict_list and m is the average length of the dictionaries in dict_list.
Auxiliary space: O(1), since the only extra space used is for the test_dict dictionary and the dict_list list, both of which are already initialized.

Method 5: Using a for loop to iterate over the dictionary list and its items and adding them to the original dictionary

Step-by-step approach:

• Initialize the original dictionary with some key-value pairs.
• Print the original dictionary to verify the initial values.
• Initialize a list of dictionaries, where each dictionary contains key-value pairs to be added to the original dictionary.
• Use a for loop to iterate over the list of dictionaries and their items, and add each key-value pair to the original dictionary.
• Print the updated dictionary to verify that the new key-value pairs have been added.

Below is the implementation of the above approach:

The original dictionary is: {'Gfg': 2, 'is': 1, 'Best': 3}
The updated dictionary: {'Gfg': 2, 'is': 1, 'Best': 3, 'for': 3, 'all': 7, 'Lazyroar': 10, 'and': 1, 'CS': 9}

Time complexity: O(N*M), where N is the length of the dictionary list and M is the average number of key-value pairs in each dictionary in the list.
Auxiliary space: O(1), as we are only modifying the original dictionary in place and not creating any additional data structures.

Method 6: Using the built-in function reduce() from the functools module and a lambda function

Step-by-step approach :

• Import the functools module.
• Define the lambda function to update the dictionary.
• Apply reduce() to the list of dictionaries. The reduce() function applies the lambda function to the test_dict and each dictionary in dict_list in turn, updating the test_dict each time.
• Print the updated dictionary.

Below is the implementation of the above approach:

The original dictionary is: {'Gfg': 2, 'is': 1, 'Best': 3}
The updated dictionary: {'Gfg': 2, 'is': 1, 'Best': 3, 'for': 3, 'all': 7, 'Lazyroar': 10, 'and': 1, 'CS': 9}

Time complexity: O(nm), where n is the length of dict_list and m is the maximum number of key-value pairs in any dictionary in dict_list.
Auxiliary space: O(1), because we are updating the original dictionary in place and not creating any new data structures.