# Python – Flatten and remove keys from Dictionary

Given a dictionary, perform flattening and removal of certain dictionary keys.

Input : test_dict = {‘a’: 14, ‘b’: 16, ‘c’: {‘d’: {‘e’: 7}}}, rem_keys = [“c”, “a”, “d”]
Output : {‘b’: 16, ‘e’: 7}
Explanation : All “c”, “a” and “d” has been removed.

Input : test_dict = {‘a’: 14, ‘b’: 16, ‘c’: {‘d’: {‘e’: 7}}}, rem_keys = [“c”, “d”, “e”]
Output : {‘a’: 14, ‘b’: 16}
Explanation : All “c”, “e” and “d” has been removed.

Method : Using recursion + isinstance() + loop

The combination of above functions can be used to solve this problem. In this, we check for dictionary as instance for nested dictionary using isinstance() and recur each time for inner dictionary. The loop is used to iterate for keys.

## Python3

 # Python3 code to demonstrate working of  # Flatten and remove keys from Dictonary # Using loop + recursion + isinstance()    # function to compute removal and flattening def hlper_fnc(test_dict, rem_keys):     if not isinstance(test_dict, dict):         return test_dict      res = {}            for key, val in test_dict.items():         rem = hlper_fnc(val, rem_keys)                    # performing removal         if key not in rem_keys:             res[key] = rem         else:             if isinstance(rem, dict):                 res.update(rem)     return res    # initializing dictionary test_dict = {'a': 14, 'b': 16, 'c': {'d': {'e': 7}}}    # printing original dictionary print("The original dictionary is : " + str(test_dict))    # initializing removal keys  rem_keys = ["c", "d"]    # calling helper function for task  res = hlper_fnc(test_dict, rem_keys)    # printing result  print("The removed and flattened dictionary : " + str(res))

Output

The original dictionary is : {‘a’: 14, ‘b’: 16, ‘c’: {‘d’: {‘e’: 7}}}
The removed and flattened dictionary : {‘a’: 14, ‘b’: 16, ‘e’: 7}

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