Given list and dictionary, extract common elements of List and Dictionary Values.
Input : test_list = [“Gfg”, “is”, “Best”, “For”], subs_dict = {4 : “Gfg”, 8 : “Geeks”, 9 : ” Good”, }
Output : [‘Gfg’]
Explanation : “Gfg” is common in both list and dictionary value.Input : test_list = [“Gfg”, “is”, “Best”, “For”, “Geeks”], subs_dict = {4 : “Gfg”, 8 : “Geeks”, 9 : ” Best”, }
Output : [‘Gfg’, “Geeks”, “Best”]
Explanation : 3 common values are extracted.
Method #1 : Using list comprehension + values()
The combination of above functionalities provide a way in which this task can be performed in a single line. In this, we extract values of dictionaries using values() and list comprehension is used to perform iteration and intersection checks.
Python3
# Python3 code to demonstrate working of # List elements and dictionary values intersection# Using list comprehension + values()# initializing listtest_list = ["Gfg", "is", "Best", "For", "Geeks"]# printing original listprint("The original list : " + str(test_list))# initializing subs. Dictionarysubs_dict = {4 : "Gfg", 8 : "Geeks", 9 : " Good", }# Intersection of elements, using "in" for checking presenceres = [ele for ele in test_list if ele in subs_dict.values()] # printing result print("Intersection elements : " + str(res)) |
The original list : ['Gfg', 'is', 'Best', 'For', 'Geeks'] Intersection elements : ['Gfg', 'Geeks']
Time Complexity: O(n), where n is the elements of dictionary
Auxiliary Space: O(n), where n is the size of dictionary
Method #2 : Using set() + intersection()
In this approach both list and dictionary values are converted to set() and then intersection is performed to get common elements.
Python3
# Python3 code to demonstrate working of # Common list elements and dictionary values # Using set() and intersection()# initializing listtest_list = ["Gfg", "is", "Best", "For", "Geeks"]# printing original listprint("The original list : " + str(test_list))# initializing subs. Dictionarysubs_dict = {4 : "Gfg", 8 : "Geeks", 9 : " Good", }# Intersection of elements, using set() to convert# intersection() for common elementsres = list(set(test_list).intersection(list(subs_dict.values()))) # printing result print("Intersection elements : " + str(res)) |
The original list : ['Gfg', 'is', 'Best', 'For', 'Geeks'] Intersection elements : ['Geeks', 'Gfg']
Time Complexity: O(n) where n is the number of elements in the list “test_list”.
Auxiliary Space: O(n) where n is the number of elements in the list “test_list”.
Method 3: Using a for loop and a conditional statement
- Create a list test_list containing some string elements.
- Create a dictionary subs_dict containing integer keys and string values.
- Create an empty list res to store the intersection elements.
- Start a for loop to iterate through each element in test_list.
- Inside the for loop, check if the current element is in subs_dict.values().
- If the current element is found in subs_dict.values(), append it to the res list.
- After the for loop is complete, print the original list test_list.
- Finally, print the list of intersection elements stored in res.
Python3
# initializing listtest_list = ["Gfg", "is", "Best", "For", "Geeks"]# initializing subs. Dictionarysubs_dict = {4 : "Gfg", 8 : "Geeks", 9 : " Good", }# initializing result listres = []# loop through test_list and check if each element is in subs_dict.values()for elem in test_list: if elem in subs_dict.values(): res.append(elem)# printing original listprint("The original list : " + str(test_list))# printing resultprint("Intersection elements : " + str(res)) |
The original list : ['Gfg', 'is', 'Best', 'For', 'Geeks'] Intersection elements : ['Gfg', 'Geeks']
Time complexity: O(n).
Auxiliary space: O(n).
