Given a list of dictionaries, remove all dictionaries which don’t have K as a value.
Examples:
Input : test_list = [{"Gfg" : 4, "is" : 8, "best" : 9}, {"Gfg" : 3, "is": 7, "best" : 5}], K = 7
Output : [{'Gfg': 4, 'is': 8, 'best': 9}]
Explanation : Resultant dictionary doesn't contain 7 as any element.
Input : test_list = [{"Gfg" : 4, "is" : 7, "best" : 9}, {"Gfg" : 3, "is": 7, "best" : 5}], K = 7
Output : []
Explanation : All contain 7 as element in List.
Method #1 : Using loop + values()
This is one of the ways in which this task can be performed. In this, we perform the task of iterating for all the dictionaries using a loop and check for value presence using values().
Python3
# Python3 code to demonstrate working of# Remove dictionary if K value not present# Using loop + values()# Initializing liststest_list = [{"Gfg": 4, "is": 8, "best": 9}, {"Gfg": 5, "is": 8, "best": 1}, {"Gfg": 3, "is": 7, "best": 6}, {"Gfg": 3, "is": 7, "best": 5}]# Printing original listprint("The original list : " + str(test_list))# initializing KK = 7res = []# Checking for K element# using loopfor sub in test_list: if K not in list(sub.values()): res.append(sub)# Printing resultprint("Filtered dictionaries : " + str(res)) |
The original list : [{'Gfg': 4, 'is': 8, 'best': 9}, {'Gfg': 5, 'is': 8, 'best': 1}, {'Gfg': 3, 'is': 7, 'best': 6}, {'Gfg': 3, 'is': 7, 'best': 5}]
Filtered dictionaries : [{'Gfg': 4, 'is': 8, 'best': 9}, {'Gfg': 5, 'is': 8, 'best': 1}]
Method #2: Using list comprehension
This is yet another way in which this task can be performed. In this, we extract all the values using one-liner using list comprehension. The values are extracted using values().
Python3
# Python3 code to demonstrate working of # Remove dictionary if K value not present# Using list comprehension# initializing liststest_list = [{"Gfg" : 4, "is" : 8, "best" : 9}, {"Gfg" : 5, "is": 8, "best" : 1}, {"Gfg" : 3, "is": 7, "best" : 6}, {"Gfg" : 3, "is": 7, "best" : 5}]# printing original listprint("The original list : " + str(test_list))# initializing K K = 7res = []# using one-liner to extract dicts with NO K value# using not in operator to check presenceres = [sub for sub in test_list if K not in list(sub.values())]# printing result print("Filtered dictionaries : " + str(res)) |
The original list : [{'Gfg': 4, 'is': 8, 'best': 9}, {'Gfg': 5, 'is': 8, 'best': 1}, {'Gfg': 3, 'is': 7, 'best': 6}, {'Gfg': 3, 'is': 7, 'best': 5}]
Filtered dictionaries : [{'Gfg': 4, 'is': 8, 'best': 9}, {'Gfg': 5, 'is': 8, 'best': 1}]
Method #3: Using filter() and lambda function:
The function takes a list of dictionaries test_list and a value K as input. If the test_list is empty, the function returns an empty list. If the first dictionary in test_list contains the value K, the function returns a list containing that dictionary concatenated with the result of calling remove_dict_without_K with the rest of the list and the same value of K. Otherwise, the function simply returns the result of calling remove_dict_without_K with the rest of the list and the same value of K.
Python3
def remove_dict_without_K(test_list, K): if not test_list: return [] elif K in test_list[0].values(): return [test_list[0]] + remove_dict_without_K(test_list[1:], K) else: return remove_dict_without_K(test_list[1:], K)# initializing liststest_list = [{"Gfg" : 4, "is" : 8, "best" : 9}, {"Gfg" : 5, "is": 8, "best" : 1}, {"Gfg" : 3, "is": 7, "best" : 6}, {"Gfg" : 3, "is": 7, "best" : 5}]# printing original listprint("The original list : " + str(test_list))# initializing K K = 7res = list(filter(lambda x: K not in x.values(), test_list))# printing result print("Filtered dictionaries : " + str(res)) |
The original list : [{'Gfg': 4, 'is': 8, 'best': 9}, {'Gfg': 5, 'is': 8, 'best': 1}, {'Gfg': 3, 'is': 7, 'best': 6}, {'Gfg': 3, 'is': 7, 'best': 5}]
Filtered dictionaries : [{'Gfg': 4, 'is': 8, 'best': 9}, {'Gfg': 5, 'is': 8, 'best': 1}]
Time complexity: O(nm), where n is the length of test_list and m is the average number of key-value pairs in each dictionary.
Auxiliary Space: O(nm), where n is the length of test_list and m is the average number of key-value pairs in each dictionary.
Method 4: Using dictionary comprehension method.
Steps:
- Initialize an empty dictionary to store the filtered dictionaries.
- Use a dictionary comprehension to iterate through each dictionary in the given list and filter out the dictionaries that have K as one of the values,
- Print the resulting dictionary of filtered dictionaries.
Python3
# Python3 code to demonstrate working of# Remove dictionary if K value not present# Using dictionary comprehension# Initializing liststest_list = [{"Gfg": 4, "is": 8, "best": 9}, {"Gfg": 5, "is": 8, "best": 1}, {"Gfg": 3, "is": 7, "best": 6}, {"Gfg": 3, "is": 7, "best": 5}]# Printing original listprint("The original list : " + str(test_list))# Initializing KK = 7# using dictionary comprehension to filter dictionariesfiltered_dict = {str(i): dict_val for i, dict_val in enumerate( test_list) if K not in dict_val.values()}# Printing resultprint("Filtered dictionaries : " + str(list(filtered_dict.values()))) |
The original list : [{'Gfg': 4, 'is': 8, 'best': 9}, {'Gfg': 5, 'is': 8, 'best': 1}, {'Gfg': 3, 'is': 7, 'best': 6}, {'Gfg': 3, 'is': 7, 'best': 5}]
Filtered dictionaries : [{'Gfg': 4, 'is': 8, 'best': 9}, {'Gfg': 5, 'is': 8, 'best': 1}]
Time complexity: O(n), where n is the number of dictionaries in the list.
Auxiliary space: O(n), as we are creating a new dictionary to store the filtered dictionaries.
