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Python – Flatten and remove keys from Dictionary
  • Last Updated : 01 Aug, 2020

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.



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# 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)) 

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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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