Python – Remove K value items from dictionary nesting

Given dictionary with multiple nestings, remove all the keys with value K.

Input : [{“Gfg” : {“a” : 5, “b” : 8, “c” : 9}}, {“is” : {“j” : 8, “k” : 10}}, {“Best” : {“i” : 16}}], K = 8 

Output : [{‘a’: 5}, {‘c’: 9}, {‘k’: 10}, {‘i’: 16}] 

Explanation : All the keys with value 8, (“b”, “j”) has been removed. 

Input : [{“Gfg” : {“a” : 5, “b” : 8, “c” : 9}}, {“is” : {“j” : 8, “k” : 10}}, {“Best” : {“i” : 16}}], K = 5 

Output : [{‘c’: 9}, {‘k’: 10}, {‘i’: 16}, {“j” : 8}, {“b” : 8}] 

Explanation : “a” with 5 as value removed.

Method #1 : Using loop + dictionary comprehension

The combination of above functions can be used to solve this problem. In this, we iterate for all the nesting keys using loop. and remove values. This approach is for single-level nesting of dictionary. 

Python3

# Python3 code to demonstrate working of
# Remove K value items from dictionary nestings
# Using dictionary comprehension + loop

# initializing list
test_list = [{"Gfg": {"a": 5, "b": 8, "c": 9}},
             {"is": {"f": 8, "j": 8, "k": 10}},
             {"Best": {"i": 16}}]

# printing original list
print("The original list is : " + str(test_list))

# initializing K
K = 8

res = list()

for sub in test_list:
    for key, val in sub.items():

        # iterating for 1st nesting only
        for key1, val1 in val.items():
            if val1 != K:
                res.append({key1: val1})

# printing result
print("The dictionary after value removal : " + str(res))

Output

The original list is : [{‘Gfg’: {‘a’: 5, ‘b’: 8, ‘c’: 9}}, {‘is’: {‘f’: 8, ‘j’: 8, ‘k’: 10}}, {‘Best’: {‘i’: 16}}] The dictionary after value removal : [{‘a’: 5}, {‘c’: 9}, {‘k’: 10}, {‘i’: 16}]

Method #2 : Using recursion + loop (For multiple nesting)

This is yet another way in which this task can be performed. In this we solve a more generic problem of catering to inner dictionary elements as well using recursion.

Python3

# Python3 code to demonstrate working of
# Remove K value items from dictionary nestings
# Using recursion + loop (For multiple nesting)

res = []

# helper function to solve problem


def hlper_fnc(test_dict):
    for key in test_dict:
        if type(test_dict[key]) == dict:
            hlper_fnc(test_dict[key])
        else:
            if test_dict[key] != K:
                res.append({key: test_dict[key]})


# initializing dictionary
test_dict = {"Gfg": {"a": 5, "b": 8, "c": 9},
             "is": {"f": 8, "l": {"j": 8, "k": 10}},
             "Best": {"i": {"k": {"o": 8}}}}

# printing original dictionary
print("The original dictionary is : " + str(test_dict))

# initializing K
K = 8

# computing result
hlper_fnc(test_dict)

# printing result
print("The dictionary after value removal : " + str(res))

Output

The original dictionary is : {‘Gfg’: {‘a’: 5, ‘b’: 8, ‘c’: 9}, ‘is’: {‘f’: 8, ‘l’: {‘j’: 8, ‘k’: 10}}, ‘Best’: {‘i’: {‘k’: {‘o’: 8}}}} The dictionary after value removal : [{‘a’: 5}, {‘c’: 9}, {‘k’: 10}]

Method #3: Using list comprehension

Step-by-step algorithm:

1. Define a list test_list of dictionaries.
2. Define a variable K that will hold the value to be removed from the nested dictionaries.
3. Define a list comprehension that iterates over each dictionary in test_list, each key-value pair in that dictionary, and each key-value pair in the nested dictionary, and filters out any nested dictionary value that is equal to K. Append each filtered key-value pair to the list res.
4. Print the resulting list of dictionaries res.

Python3

test_list = [{"Gfg": {"a": 5, "b": 8, "c": 9}}, {
    "is": {"f": 8, "j": 8, "k": 10}}, {"Best": {"i": 16}}]

K = 8

# printing original dictionary
print("The original dictionary is : " + str(test_list))

# Using list comprehension to filter out values that are equal to K
res = [{key1: val1} for sub in test_list for key, val in sub.items()
       for key1, val1 in val.items() if val1 != K]

# printing the resulting list of list
print("The dictionary after value removal: " + str(res))

Output

The original dictionary is : [{'Gfg': {'a': 5, 'b': 8, 'c': 9}}, {'is': {'f': 8, 'j': 8, 'k': 10}}, {'Best': {'i': 16}}]
The dictionary after value removal: [{'a': 5}, {'c': 9}, {'k': 10}, {'i': 16}]

Time Complexity: O(nm), where n is the number of dictionaries in test_list and m is the total number of key-value pairs in all the nested dictionaries.
Auxiliary Space: O(nm), where n is the number of dictionaries in test_list and m is the total number of key-value pairs in all the nested dictionaries.