Given a dictionary, extract all the items that match given values in the list
Example:
Input : test_dict = {“Gfg” : 3, “is” : 5, “for” : 8, “Geeks” : 10, “Best” : 16 }, sub_list = [5, 4, 10, 20, 16, 8]
Output : {‘is’: 5, ‘Geeks’: 10, ‘Best’: 16, “for” : 8}
Explanation : All values matching list values extracted along with keys.Input : test_dict = {“Gfg” : 3, “is” : 5, “for” : 8, “Geeks” : 10, “Best” : 16 }, sub_list = [5, 4, 10]
Output : {‘is’: 5, ‘Geeks’: 10}
Explanation : All values matching list values extracted along with keys.
Method 1: Using a loop
Use of a ‘for loop’ is one of the ways in which this task can be performed. In this, we iterate for all the keys and check if the value is present in a custom list, if yes, then it’s returned.
Python3
# initializing dictionarytest_dict = {"Gfg": 3, "is": 5, "for": 8, "Geeks": 10, "Best": 16}# printing original dictionaryprint("The original dictionary is : " + str(test_dict))# initializing listsub_list = [5, 4, 10, 20, 16]# Using loop to perform iterationres = dict()for key in test_dict: if test_dict[key] in sub_list: res[key] = test_dict[key]# printing resultprint("Extracted items : " + str(res)) |
Output:
The original dictionary is : {‘Gfg’: 3, ‘is’: 5, ‘for’: 8, ‘Geeks’: 10, ‘Best’: 16}
Extracted items : {‘is’: 5, ‘Geeks’: 10, ‘Best’: 16}
Method 2: Using dictionary comprehension
This is a one-liner with the help of which this task can be performed. In this, we iterate for all keys using dictionary comprehension in one-liner approach.
Python3
# initializing dictionarytest_dict = {"Gfg": 3, "is": 5, "for": 8, "Geeks": 10, "Best": 16}# printing original dictionaryprint("The original dictionary is : " + str(test_dict))# initializing listsub_list = [5, 4, 10, 20, 16]# dictionary comprehension to compile logic in one dictionary# in operator used to check value existenceres = {key: val for key, val in test_dict.items() if val in sub_list}# printing resultprint("Extracted items : " + str(res)) |
Output:
The original dictionary is : {‘Gfg’: 3, ‘is’: 5, ‘for’: 8, ‘Geeks’: 10, ‘Best’: 16}
Extracted items : {‘is’: 5, ‘Geeks’: 10, ‘Best’: 16}
Method 3: Using set intersection
Approach
we can convert both the dictionary values and the given list into sets, and then perform a set intersection operation to get the common values. Finally, we can create a new dictionary containing the items with the common values.
Algorithm
1. Convert the dictionary values into a set using the set() function.
2. Convert the given list into a set using the set() function.
3. Perform a set intersection operation on the two sets using the & operator.
4. Create a new dictionary containing the items with the common values.
5. Return the new dictionary.
Python3
def extract_items(test_dict, sub_list): dict_values_set = set(test_dict.values()) sub_list_set = set(sub_list) common_values = dict_values_set & sub_list_set filtered_dict = {k: v for k, v in test_dict.items() if v in common_values} return filtered_dicttest_dict = {"Gfg": 3, "is": 5, "for": 8, "Geeks": 10, "Best": 16}sub_list = [5, 4, 10, 20, 16]print(extract_items(test_dict, sub_list)) |
{'is': 5, 'Geeks': 10, 'Best': 16}
Time Complexity: O(N), where N is the number of items in the dictionary.
Space Complexity: O(N), where N is the number of items in the dictionary.
Method 4 : using the filter() function along with a lambda function.
The steps involved in this approach are as follows:
Initialize the dictionary test_dict and list sub_list.
Use the filter() function along with a lambda function to filter the items from the dictionary test_dict based on whether the value is present in the sub_list or not.
Convert the filtered items into a dictionary using the dict() constructor.
Print the extracted items.
Python3
# initializing dictionarytest_dict = {"Gfg": 3, "is": 5, "for": 8, "Geeks": 10, "Best": 16}# printing original dictionaryprint("The original dictionary is : " + str(test_dict))# initializing listsub_list = [5, 4, 10, 20, 16]# Using filter() function to extract itemsres = dict(filter(lambda item: item[1] in sub_list, test_dict.items()))# printing resultprint("Extracted items : " + str(res)) |
The original dictionary is : {'Gfg': 3, 'is': 5, 'for': 8, 'Geeks': 10, 'Best': 16}
Extracted items : {'is': 5, 'Geeks': 10, 'Best': 16}
The time complexity of this approach is O(n), where n is the number of items in the dictionary test_dict.
The auxiliary space complexity is also O(n), as a new dictionary is created to store the filtered items.
