Python - Value Dictionary from Record List

Sometimes, while working with Python Records lists, we can have problems in which, we need to reform the dictionary taking just values of the binary dictionary. This can have applications in many domains which work with data. Let us discuss certain ways in which this task can be performed. 

Method #1 : Using loop + values() + update()

The combination of the above functions can be used to perform this task. In this the values are extracted using values() and updating the new dictionary is done using update(). 

The original list is : [{1: 'gfg', 2: 'best'}, {3: 'for', 4: 'geeks'}]
The values dictionary is : {'gfg': 'best', 'for': 'geeks'}

Time Complexity: O(n) where n is the total number of values in the list “test_list”. 
Auxiliary Space: O(n) where n is the total number of values in the list “test_list”. 

Method #2 : Using zip() + iter()

The combination of the above functions can also be used to perform this task. In this, we perform to convert the list to iterator and pairing of values is done using zip(). 

The original list is : [{1: 'gfg', 2: 'best'}, {3: 'for', 4: 'geeks'}]
The values dictionary is : {'gfg': 'best', 'for': 'geeks'}

Time Complexity: O(n) where n is the number of elements in the dictionary. The zip() + iter() is used to perform the task and it takes O(n) time.
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the dictionary.

Method #3:Using List comprehension

Algorithm

1. Initialize an empty dictionary res.
2. Iterate over each dictionary sub in the record list test_list.
3. Get the values of the dictionary sub using the values() method, and update the res dictionary with these values using the update() method.
4. Convert the resulting res dictionary to a standard dictionary using the dict() function.
5. Print the resulting dictionary.

The original list is : [{1: 'gfg', 2: 'best'}, {3: 'for', 4: 'geeks'}]
The values dictionary is : {'gfg': 'best', 'for': 'geeks'}

Time Complexity: O(N*M)

The loop iterates over each dictionary in the record list once. Therefore, the time complexity of this algorithm is O(N*M), where N is the number of dictionaries in the list and M is the average size of each dictionary.
Auxiliary Space: O(N*M)

The space complexity of this algorithm is O(NM), where N is the number of dictionaries in the list and M is the average size of each dictionary. The res dictionary stores all the values from each dictionary in the list, which can take up NM space. Additionally, other variables and temporary memory used during the execution of the algorithm can also contribute to space complexity.

Method #4: Using for loop+values() method

Approach:

1. Create an empty dictionary.
2. Initiate a for loop to traverse list of dictionaries.
3. Extract values of each dictionary using values() method.
4. Make 0th index value of values list as key and 1st index value of values list as value.
5. Display the dictionary.

Below is the implementation of the above approach:

The original list is : [{1: 'gfg', 2: 'best'}, {3: 'for', 4: 'geeks'}]
The values dictionary is : {'gfg': 'best', 'for': 'geeks'}

Time Complexity: O(M*N), where M is the length of list of dictionaries and N is the length of each dictionary
Auxiliary Space: O(M*N), where M is the number of keys and N is the number of values

Method #5: Using dictionary comprehension

The original list is : [{1: 'gfg', 2: 'best'}, {3: 'for', 4: 'geeks'}]
The values dictionary is : {'gfg': 'best', 'for': 'geeks'}

Time complexity: O(n), where n is the length of test_list.
Auxiliary space: O(n)

Method #6: Using map() and lambda

The original list is : [{1: 'gfg', 2: 'best'}, {3: 'for', 4: 'geeks'}]
The values dictionary is : {'gfg': 'best', 'for': 'geeks'}

Time complexity: O(n), where n is the length of test_list.
Auxiliary space: O(n)