Given list of dictionaries, remove dictionary whose Key K is N.
Input : test_list = [{“Gfg” : 3, “is” : 7, “Best” : 8}, {“Gfg” : 9, “is” : 2, “Best” : 9}, {“Gfg” : 5, “is” : 4, “Best” : 10}, {“Gfg” : 3, “is” : 6, “Best” : 15}], K = “Gfg”, N = 9
Output : [{“Gfg” : 3, “is” : 7, “Best” : 8}, {“Gfg” : 5, “is” : 4, “Best” : 10}, {“Gfg” : 3, “is” : 6, “Best” : 15}]
Explanation : All elements are extracted which have “Gfg” other than 9.Input : test_list = [{“Gfg” : 3, “is” : 7, “Best” : 8}, {“Gfg” : 9, “is” : 2, “Best” : 9}, {“Gfg” : 5, “is” : 4, “Best” : 10}, {“Gfg” : 3, “is” : 6, “Best” : 15}], K = “Best”, N = 10
Output : [{“Gfg” : 3, “is” : 7, “Best” : 8}, {“Gfg” : 9, “is” : 2, “Best” : 9}, {“Gfg” : 3, “is” : 6, “Best” : 15}]
Explanation : All elements are extracted which have “Best” other than 10.
Method #1 : Using list comprehension
This is one of the ways in which this task can be performed. In this, we extract and iterate using conditional checks using list comprehension in one liner.
Python3
# Python3 code to demonstrate working of # Remove Dictionaries whose Key(K) is N# Using list comprehension # initializing listtest_list = [{"Gfg" : 3, "is" : 7, "Best" : 8}, {"Gfg" : 9, "is" : 2, "Best" : 9}, {"Gfg" : 5, "is" : 4, "Best" : 10}, {"Gfg" : 3, "is" : 6, "Best" : 8}] # printing original listprint("The original list : " + str(test_list)) # initializing K K = "Gfg" # initializing N N = 5 # returning only dictionaries with "Gfg" key not 5 res = [sub for sub in test_list if sub[K] != N] # printing result print("The extracted dictionaries : " + str(res)) |
The original list : [{'Gfg': 3, 'is': 7, 'Best': 8}, {'Gfg': 9, 'is': 2, 'Best': 9}, {'Gfg': 5, 'is': 4, 'Best': 10}, {'Gfg': 3, 'is': 6, 'Best': 8}]
The extracted dictionaries : [{'Gfg': 3, 'is': 7, 'Best': 8}, {'Gfg': 9, 'is': 2, 'Best': 9}, {'Gfg': 3, 'is': 6, 'Best': 8}]
Method #2 : Using filter() + lambda
This is yet another way in which this task can be performed. In this, we use conditionals using filter() and lambda function is for checking for N value.
Python3
# Python3 code to demonstrate working of # Remove Dictionaries whose Key(K) is N# Using filter() + lambda # initializing listtest_list = [{"Gfg" : 3, "is" : 7, "Best" : 8}, {"Gfg" : 9, "is" : 2, "Best" : 9}, {"Gfg" : 5, "is" : 4, "Best" : 10}, {"Gfg" : 3, "is" : 6, "Best" : 8}] # printing original listprint("The original list : " + str(test_list)) # initializing K K = "Gfg" # initializing N N = 5 # Using filter() to check for N valueres = list(filter(lambda x: x[K] != N, test_list)) # printing result print("The extracted dictionaries : " + str(res)) |
The original list : [{'Gfg': 3, 'is': 7, 'Best': 8}, {'Gfg': 9, 'is': 2, 'Best': 9}, {'Gfg': 5, 'is': 4, 'Best': 10}, {'Gfg': 3, 'is': 6, 'Best': 8}]
The extracted dictionaries : [{'Gfg': 3, 'is': 7, 'Best': 8}, {'Gfg': 9, 'is': 2, 'Best': 9}, {'Gfg': 3, 'is': 6, 'Best': 8}]
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