Python – Extract Key’s Value, if Key Present in List and Dictionary

Given a list, dictionary, and a Key K, print the value of K from the dictionary if the key is present in both, the list and the dictionary.

Input : test_list = ["Gfg", "is", "Good", "for", "Geeks"], 
                test_dict = {"Gfg" : 5, "Best" : 6}, K = "Gfg" 
Output : 5 
Explanation : "Gfg" is present in list and has value 5 in dictionary. 

Input : test_list = ["Good", "for", "Geeks"], 
                test_dict = {"Gfg" : 5, "Best" : 6}, K = "Gfg" 
Output : None 
Explanation : "Gfg" not present in List.

Method #1 : Using all() + generator expression

The combination of the above functions offers one of the ways in which this problem can be solved. In this, we use all() to check for occurrence in both dictionary and list. If the result is true value is extracted from to result.

The original list : ['Gfg', 'is', 'Good', 'for', 'Geeks']
The original Dictionary : {'Gfg': 2, 'is': 4, 'Best': 6}
Extracted Value : 2

Time complexity: O(n), where n is the total number of elements in the test_list. This is because, in the worst case, each element of the list needs to be checked for its presence in both the list and the dictionary, resulting in a time complexity of O(n).
Auxiliary space: O(1), as it only requires a constant amount of additional memory to store the result and temporary variables.

Method #2 : Using set() + intersection()

This is another way to check for the key’s presence in both containers. In this, we compute the intersection of all values of list and dict keys and test for the Key’s occurrence in that.

The original list : ['Gfg', 'is', 'Good', 'for', 'Geeks']
The original Dictionary : {'Gfg': 2, 'is': 4, 'Best': 6}
Extracted Value : 2

Time complexity: O(n), where n is the length of the list.
Auxiliary space: O(m), where m is the number of unique keys in the dictionary that match the elements in the list.

Method #3: Using in operator

The original list : ['Gfg', 'is', 'Good', 'for', 'Geeks']
The original Dictionary : {'Gfg': 2, 'is': 4, 'Best': 6}
Extracted Value : 2

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

Method 4: Using operator.countOf() method

The original list : ['Gfg', 'is', 'Good', 'for', 'Geeks']
The original Dictionary : {'Gfg': 2, 'is': 4, 'Best': 6}
Extracted Value : 2

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

Method 5: Using any() + dictionary.get() method

Step-by-step approach:

• Use the any() function to check if the string variable K is present in both the test_list and test_dict.
• If K is present in both test_list and test_dict, use the get() method of the dictionary to retrieve the value corresponding to the key K and store it in a variable named res.
• Print the extracted value of K from the dictionary in a human-readable format by converting it to a string using the str() function, and concatenating it with a message string.

Below is the implementation of the above approach:

The original list : ['Gfg', 'is', 'Good', 'for', 'Geeks']
The original Dictionary : {'Gfg': 2, 'is': 4, 'Best': 6}
Extracted Value : 2

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

Method #6: Using try-except block to handle KeyErrors

Use a try-except block to handle the KeyError that may occur if K is not present in test_dict. First, check if K is present in both test_list and test_dict.keys(). If yes, extract the value of K from test_dict. If no, set the result to None. If a KeyError occurs, handle it by setting the result to None. Finally, print the result.

Extracted Value : 2

Time complexity: O(n), where n is the length of the test_list. This is because we check if K is present in both test_list and test_dict.keys(), which takes O(n) time in the worst case.
Auxiliary space: O(1), because we use a constant amount of extra space to store the variables test_list, test_dict, K, and res, and we don’t create any additional data structures in the code.

Method #7: Using try-except block with .get() method

This method involves using a try-except block to check if the key exists in the dictionary and retrieve its value using the .get() method.

Follow the below steps to implement the above idea:

• Initialize the list, dictionary, and key, same as in the original code.
• Use a try-except block to handle the KeyError exception that occurs if the key is not found in the dictionary.
• Inside the try block, retrieve the value of the key using the .get() method on the dictionary. If the key is not found, .get() will return None by default.
• If the key is found in both the list and dictionary, return the value of the key. If not, return None.

Below is the implementation of the above approach:

The original list : ['Gfg', 'is', 'Good', 'for', 'Geeks']
The original Dictionary : {'Gfg': 2, 'is': 4, 'Best': 6}
Extracted Value : 2

Time complexity: O(n), Where n is the length of test_list.
Auxiliary space: O(1) for storing the result and constant extra space for the try-except block.

Method #8: Using a for loop to iterate through the list and dictionary

Step-by-step approach:

• Initialize the list, dictionary, and key as given in the problem statement:
• Use a for loop to iterate through the list and dictionary:
  • Checks each item in the list. If an item is equal to the key, it uses the get() method to retrieve the corresponding value from the dictionary. 
  • If the key is not found, res is set to None. The loop stops as soon as a matching item is found.
• Print the extracted value:

Method 9: Extracted Value : 2

Time complexity: O(n), where n is the length of the list
Auxiliary space: O(1)

Method #9: Using numpy:

Algorithm:

1. Import numpy module
2. Initialize list, dictionary and key
3. Use numpy’s isin() function to check if the key is present in both the list and dictionary
4. If the key is present in both the list and dictionary, extract the value from the dictionary
5. Print the extracted value

Output:

The original list : ['Gfg', 'is', 'Good', 'for', 'Geeks']
The original Dictionary : {'Gfg': 2, 'is': 4, 'Best': 6}
Extracted Value : 2

Time Complexity: O(n), where n is the length of the list and the number of keys in the dictionary. Both the isin() function and dictionary keys() function have a time complexity of O(n).
Auxiliary Space: O(n), where n is the length of the list and the number of keys in the dictionary. The isin() function creates a numpy array of length n, and the dictionary takes up space proportional to the number of keys.

Method#10: Using Recursive method.

Step-by-step approach:

• Define a function extract_value that takes in three arguments: test_list, test_dict, and K.
• Check if test_list is empty.
• If test_list is empty, return None.
• Otherwise, check if the first element of test_list is equal to K.
• If the first element of test_list is equal to K, return the value associated with key K in test_dict using the get method.
• Otherwise, call extract_value recursively with the remaining elements of test_list.

Below is the implementation of the above approach:

Extracted Value : 2

Time complexity: O(n), where n is the length of test_list. This is because we need to iterate over all elements of test_list to find the key K.
Auxiliary space: O(n), where n is the length of test_list. This is because we need to store n recursive calls on the call stack.

manjeet_04

Follow

Improve