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Учитывая сплющенный словарь, задача состоит в том, чтобы преобразовать этот словарь во вложенный словарь, где ключи необходимо разделить на»_», учитывая, где будет запущен вложенный словарь.

Метод № 1: Использование наивного подхода 

# Python code to demonstrate
# conversion 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 hierarchical flow
	lists = ([*k.split("_"), v] for k, v in dct.items())

	# insert each list into the result
	for lst in lists:
		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}

# printing 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)))

Выход:

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}}}

Метод № 2: Использование диктанта по умолчанию и рекурсивного подхода

# Python code to demonstrate
# conversion of flattened dictionary
# into nested dictionary

# code to convert 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}

# printing initial dictionary
print ("initial_dictionary", str(ini_dict))

# printing final dictionary
print ("final_dictionary", str(nest_dict(ini_dict)))

Выход:

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}}}

Метод № 3: Использование reduce и getitem

# Python code to demonstrate
# conversion 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()

# converting 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}

# printing 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)))

Выход:

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}}}