Given a list of dictionaries, the task is to write a Python program to filter dictionaries on basis of elements of Kth key in the list.
Examples:
Input : test_list = [{"Gfg" : 3, "is" : 5, "best" : 10},
{"Gfg" : 5, "is" : 1, "best" : 1},
{"Gfg" : 8, "is" : 3, "best" : 9},
{"Gfg" : 9, "is" : 9, "best" : 8},
{"Gfg" : 4, "is" : 10, "best" : 7}], K = "best", search_list = [1, 9, 8, 4, 5]
Output : [{'Gfg': 5, 'is': 1, 'best': 1}, {'Gfg': 8, 'is': 3, 'best': 9}, {'Gfg': 9, 'is': 9, 'best': 8}]
Explanation : Dictionaries with "best" as key and values other than 1, 9, 8, 4, 5 are omitted.
Input : test_list = [{"Gfg" : 3, "is" : 5, "best" : 10},
{"Gfg" : 5, "is" : 1, "best" : 1},
{"Gfg" : 8, "is" : 3, "best" : 9}], K = "best", search_list = [1, 9, 4, 5]
Output : [{'Gfg': 5, 'is': 1, 'best': 1}, {'Gfg': 8, 'is': 3, 'best': 9}]
Explanation : Dictionaries with "best" as key and values other than 1, 9, 4, 5 are omitted.
Method #1 : Using loop + conditional statements
In this, key-value pairs are added to the resultant dictionary after checking for Kth keys values in the list using conditionals, iterated using a loop.
Python3
# Python3 code to demonstrate working of# Filter dictionaries by values in Kth Key in list# Using loop + conditional statements# initializing listtest_list = [{"Gfg": 3, "is": 5, "best": 10}, {"Gfg": 5, "is": 1, "best": 1}, {"Gfg": 8, "is": 3, "best": 9}, {"Gfg": 9, "is": 9, "best": 8}, {"Gfg": 4, "is": 10, "best": 7}]# printing original listprint("The original list is : " + str(test_list))# initializing search_listsearch_list = [1, 9, 8, 4, 5]# initializing KK = "best"res = []for sub in test_list: # checking if Kth key's value present in search_list if sub[K] in search_list: res.append(sub)# printing resultprint("Filtered dictionaries : " + str(res)) |
The original list is : [{'Gfg': 3, 'is': 5, 'best': 10}, {'Gfg': 5, 'is': 1, 'best': 1}, {'Gfg': 8, 'is': 3, 'best': 9}, {'Gfg': 9, 'is': 9, 'best': 8}, {'Gfg': 4, 'is': 10, 'best': 7}]
Filtered dictionaries : [{'Gfg': 5, 'is': 1, 'best': 1}, {'Gfg': 8, 'is': 3, 'best': 9}, {'Gfg': 9, 'is': 9, 'best': 8}]
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #2: Using list comprehension
Similar to the above method, list comprehension is used to provide shorthand to the method used above.
Python3
# Python3 code to demonstrate working of# Filter dictionaries by values in Kth Key in list# Using list comprehension# initializing listtest_list = [{"Gfg": 3, "is": 5, "best": 10}, {"Gfg": 5, "is": 1, "best": 1}, {"Gfg": 8, "is": 3, "best": 9}, {"Gfg": 9, "is": 9, "best": 8}, {"Gfg": 4, "is": 10, "best": 7}, ]# printing original listprint("The original list is : " + str(test_list))# initializing search_listsearch_list = [1, 9, 8, 4, 5]# initializing KK = "best"# list comprehension as shorthand for solving problemres = [sub for sub in test_list if sub[K] in search_list]# printing resultprint("Filtered dictionaries : " + str(res)) |
The original list is : [{'Gfg': 3, 'is': 5, 'best': 10}, {'Gfg': 5, 'is': 1, 'best': 1}, {'Gfg': 8, 'is': 3, 'best': 9}, {'Gfg': 9, 'is': 9, 'best': 8}, {'Gfg': 4, 'is': 10, 'best': 7}]
Filtered dictionaries : [{'Gfg': 5, 'is': 1, 'best': 1}, {'Gfg': 8, 'is': 3, 'best': 9}, {'Gfg': 9, 'is': 9, 'best': 8}]
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 3: Using filter() + lambda function
Python3
# Python3 code to demonstrate working of# Filter dictionaries by values in Kth Key in list# Using filter() + lambda function# initializing listtest_list = [{"Gfg": 3, "is": 5, "best": 10}, {"Gfg": 5, "is": 1, "best": 1}, {"Gfg": 8, "is": 3, "best": 9}, {"Gfg": 9, "is": 9, "best": 8}, {"Gfg": 4, "is": 10, "best": 7}]# printing original listprint("The original list is : " + str(test_list))# initializing search_listsearch_list = [1, 9, 8, 4, 5]# initializing KK = "best"# using filter() and lambda function to filter dictionariesres = list(filter(lambda sub: sub[K] in search_list, test_list))# printing resultprint("Filtered dictionaries : " + str(res)) |
The original list is : [{'Gfg': 3, 'is': 5, 'best': 10}, {'Gfg': 5, 'is': 1, 'best': 1}, {'Gfg': 8, 'is': 3, 'best': 9}, {'Gfg': 9, 'is': 9, 'best': 8}, {'Gfg': 4, 'is': 10, 'best': 7}]
Filtered dictionaries : [{'Gfg': 5, 'is': 1, 'best': 1}, {'Gfg': 8, 'is': 3, 'best': 9}, {'Gfg': 9, 'is': 9, 'best': 8}]
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #4: Using a for loop and a temporary list
Using a for loop and a temporary list to store the filtered dictionaries.
Python3
# Python3 code to demonstrate working of# Filter dictionaries by values in Kth Key in list# Using for loop and temporary list# initializing listtest_list = [{"Gfg": 3, "is": 5, "best": 10}, {"Gfg": 5, "is": 1, "best": 1}, {"Gfg": 8, "is": 3, "best": 9}, {"Gfg": 9, "is": 9, "best": 8}, {"Gfg": 4, "is": 10, "best": 7}]# initializing search_listsearch_list = [1, 9, 8, 4, 5]# initializing KK = "best"# using for loop and temporary list to filter dictionariestemp_list = []for d in test_list: if d[K] in search_list: temp_list.append(d)# storing filtered dictionaries in resres = temp_list# printing resultprint("Filtered dictionaries : " + str(res)) |
Filtered dictionaries : [{'Gfg': 5, 'is': 1, 'best': 1}, {'Gfg': 8, 'is': 3, 'best': 9}, {'Gfg': 9, 'is': 9, 'best': 8}]
Time complexity: O(nk), where n is the number of dictionaries in the list and k is the number of key-value pairs in each dictionary
Auxiliary space: O(nk), where n is the number of dictionaries in the list and k is the number of key-value pairs in each dictionary
Method #5: Using dictionary comprehension
- Initializing list
- printing original list
- Initializing search_list
- Initializing K
- Using dictionary comprehension to filter dictionaries
- printing result
Python3
# Python3 code to demonstrate working of# Filter dictionaries by values in Kth Key in list# Using dictionary comprehension# initializing listtest_list = [{"Gfg": 3, "is": 5, "best": 10}, {"Gfg": 5, "is": 1, "best": 1}, {"Gfg": 8, "is": 3, "best": 9}, {"Gfg": 9, "is": 9, "best": 8}, {"Gfg": 4, "is": 10, "best": 7}]# printing original listprint("The original list is : " + str(test_list))# initializing search_listsearch_list = [1, 9, 8, 4, 5]# initializing KK = "best"# using dictionary comprehension to filter dictionariesres = [sub for sub in test_list if sub.get(K) in search_list]# printing resultprint("Filtered dictionaries : " + str(res)) |
The original list is : [{'Gfg': 3, 'is': 5, 'best': 10}, {'Gfg': 5, 'is': 1, 'best': 1}, {'Gfg': 8, 'is': 3, 'best': 9}, {'Gfg': 9, 'is': 9, 'best': 8}, {'Gfg': 4, 'is': 10, 'best': 7}]
Filtered dictionaries : [{'Gfg': 5, 'is': 1, 'best': 1}, {'Gfg': 8, 'is': 3, 'best': 9}, {'Gfg': 9, 'is': 9, 'best': 8}]
Time complexity: O(n), where n is the length of the input list test_list.
Auxiliary space: O(k), where k is the number of filtered dictionaries (the length of the output list res).
Method #6: Using map() and lambda function
- Define a lambda function that takes a dictionary d and returns d[K] (the value of the Kth key).
- Use map() to apply the lambda function to each dictionary in test_list and store the result in a list.
- Use filter() to filter the dictionaries in the list based on whether d[K] is in search_list.
Python3
# Python3 code to demonstrate working of# Filter dictionaries by values in Kth Key in list# Using map() + lambda function + filter()# initializing listtest_list = [{"Gfg": 3, "is": 5, "best": 10}, {"Gfg": 5, "is": 1, "best": 1}, {"Gfg": 8, "is": 3, "best": 9}, {"Gfg": 9, "is": 9, "best": 8}, {"Gfg": 4, "is": 10, "best": 7}]# printing original listprint("The original list is : " + str(test_list))# initializing search_listsearch_list = [1, 9, 8, 4, 5]# initializing KK = "best"# using map() + lambda function to extract Kth key valuevalues = list(map(lambda d: d[K], test_list))# using filter() to filter dictionaries based on valuesres = list(filter(lambda d: d[K] in search_list, test_list))# printing resultprint("Filtered dictionaries : " + str(res)) |
The original list is : [{'Gfg': 3, 'is': 5, 'best': 10}, {'Gfg': 5, 'is': 1, 'best': 1}, {'Gfg': 8, 'is': 3, 'best': 9}, {'Gfg': 9, 'is': 9, 'best': 8}, {'Gfg': 4, 'is': 10, 'best': 7}]
Filtered dictionaries : [{'Gfg': 5, 'is': 1, 'best': 1}, {'Gfg': 8, 'is': 3, 'best': 9}, {'Gfg': 9, 'is': 9, 'best': 8}]
The time complexity of this approach is O(N), where N is the number of dictionaries in test_list, since both map() and filter() loop through the list once. The auxiliary space is O(N), since we create a list of length N to store the extracted values of the Kth key.
Method 7: Using list comprehension + any()
Initialize an empty list “res” to store the filtered dictionaries.
Use a list comprehension to iterate over “test_list” and check if the value of key “K” is present in “search_list” using the “any()” function.
If the condition is True, then append the current dictionary to the “res” list.
Print the filtered list of dictionaries.
Python3
# Python3 code to demonstrate working of# Filter dictionaries by values in Kth Key in list# Using list comprehension and the any() function# initializing listtest_list = [{"Gfg": 3, "is": 5, "best": 10}, {"Gfg": 5, "is": 1, "best": 1}, {"Gfg": 8, "is": 3, "best": 9}, {"Gfg": 9, "is": 9, "best": 8}, {"Gfg": 4, "is": 10, "best": 7}]# printing original listprint("The original list is : " + str(test_list))# initializing search_listsearch_list = [1, 9, 8, 4, 5]# initializing KK = "best"# using list comprehension and any() function to filter dictionaries based on valuesres = [d for d in test_list if d[K] in search_list]# printing resultprint("Filtered dictionaries : " + str(res)) |
The original list is : [{'Gfg': 3, 'is': 5, 'best': 10}, {'Gfg': 5, 'is': 1, 'best': 1}, {'Gfg': 8, 'is': 3, 'best': 9}, {'Gfg': 9, 'is': 9, 'best': 8}, {'Gfg': 4, 'is': 10, 'best': 7}]
Filtered dictionaries : [{'Gfg': 5, 'is': 1, 'best': 1}, {'Gfg': 8, 'is': 3, 'best': 9}, {'Gfg': 9, 'is': 9, 'best': 8}]
Time complexity: O(n), where “n” is the length of the input list “test_list”. This is because we are iterating over the list once using the list comprehension.
Auxiliary Space: O(k), where “k” is the length of the output list “res”. This is because we are storing the filtered dictionaries in a new list “res”.
