Python | Convert flattened dictionary into nested dictionary

Given a flattened dictionary, the task is to convert that dictionary into a nested dictionary where keys are needed to be split at ‘_’ considering where nested dictionary will be started.

Method #1: Using Naive Approach

# Python code to demonstrate 
# conersion of flattened dictionary 
# into nested dictionary 
  
  
def insert(dct, lst): 
    for x in lst[:-2]: 
        dct[x] = dct = dct.get(x, dict()) 
    dct.update({lst[-2]: lst[-1]}) 
      
  
def convert_nested(dct): 
    # empty dict to store the result 
    result = dict() 
  
    # create an iterator of lists  
    # representing nested or hierarchial flow 
    lsts = ([*k.split("_"), v] for k, v in dct.items()) 
  
    # insert each list into the result 
    for lst in lsts: 
        insert(result, lst) 
    return result 
          
# initialising_dictionary 
ini_dict = {'Geeks_for_for':1,'Geeks_for_geeks':4, 
            'for_geeks_Geeks':3,'geeks_Geeks_for':7} 
  
# priniting initial dictionary 
print ("initial_dictionary", str(ini_dict)) 
  
# code to convert ini_dict to nested  
# dictionary splitting_dict_keys 
_split_dict = [[*a.split('_'), b] for a, b in ini_dict.items()] 
  
  
# printing final dictionary 
print ("final_dictionary", str(convert_nested(ini_dict))) 

Output:

initial_dictionary {‘for_geeks_Geeks’: 3, ‘Geeks_for_geeks’: 4, ‘geeks_Geeks_for’: 7, ‘Geeks_for_for’: 1}
final_dictionary {‘Geeks’: {‘for’: {‘for’: 1, ‘geeks’: 4}}, ‘for’: {‘geeks’: {‘Geeks’: 3}}, ‘geeks’: {‘Geeks’: {‘for’: 7}}}

Method #2: Using default dict and recursive approach

# Python code to demonstrate 
# conersion of flattened dictionary 
# into nested dictionary 
  
# code to conert dict into nested dict 
def nest_dict(dict1): 
    result = {} 
    for k, v in dict1.items(): 
          
        # for each key call method split_rec which 
        # will split keys to form recursively  
        # nested dictionary 
        split_rec(k, v, result) 
    return result 
  
def split_rec(k, v, out): 
      
    # splitting keys in dict 
    # calling_recursively to break items on '_' 
    k, *rest = k.split('_', 1) 
    if rest: 
        split_rec(rest[0], v, out.setdefault(k, {})) 
    else: 
        out[k] = v 
  
          
# initialising_dictionary 
ini_dict = {'Geeks_for_for':1,'Geeks_for_geeks':4, 
            'for_geeks_Geeks':3,'geeks_Geeks_for':7} 
  
# priniting initial dictionary 
print ("initial_dictionary", str(ini_dict)) 
  
# printing final dictionary 
print ("final_dictionary", str(nest_dict(ini_dict))) 

Output:

initial_dictionary {‘for_geeks_Geeks’: 3, ‘Geeks_for_for’: 1, ‘Geeks_for_geeks’: 4, ‘geeks_Geeks_for’: 7}
final_dictionary {‘for’: {‘geeks’: {‘Geeks’: 3}}, ‘geeks’: {‘Geeks’: {‘for’: 7}}, ‘Geeks’: {‘for’: {‘for’: 1, ‘geeks’: 4}}}

Method #3: Using reduce and getitem

# Python code to demonstrate 
# conersion of flattened dictionary 
# into nested dictionary 
  
from collections import defaultdict 
from functools import reduce
from operator import getitem 
  
  
def getFromDict(dataDict, mapList): 
      
    # Iterate nested dictionary 
    return reduce(getitem, mapList, dataDict) 
      
# instantiate nested defaultdict of defaultdicts 
tree = lambda: defaultdict(tree) 
d = tree() 
  
# conerting default_dict_to regular dict 
def default_to_regular(d): 
      
    """Convert nested defaultdict to regular dict of dicts."""
    if isinstance(d, defaultdict): 
        d = {k: default_to_regular(v) for k, v in d.items()} 
    return d 
          
# initialising_dictionary 
ini_dict = {'Geeks_for_for':1,'Geeks_for_geeks':4, 
            'for_geeks_Geeks':3,'geeks_Geeks_for':7} 
  
# priniting initial dictionary 
print ("initial_dictionary", str(ini_dict)) 
  
  
# code to convert ini_dict to nested dictionary 
# iterating_over_dict 
for k, v in ini_dict.items(): 
      
    # splitting keys 
    * keys, final_key = k.split('_') 
    getFromDict(d, keys)[final_key] = v 
  
# printing final dictionary 
print ("final_dictionary", str(default_to_regular(d))) 

Output:

initial_dictionary {‘Geeks_for_for’: 1, ‘Geeks_for_geeks’: 4, ‘for_geeks_Geeks’: 3, ‘geeks_Geeks_for’: 7}
final_dictionary {‘Geeks’: {‘for’: {‘geeks’: 4, ‘for’: 1}}, ‘geeks’: {‘Geeks’: {‘for’: 7}}, ‘for’: {‘geeks’: {‘Geeks’: 3}}}

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