Dado un diccionario aplanado, la tarea es convertir ese diccionario en un diccionario anidado donde las claves deben dividirse en ‘_’ considerando dónde se iniciará el diccionario anidado.
Método #1: Usar un enfoque ingenuo
Python3
# 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)))
Salida:
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}}, ‘para’: {‘geeks’: {‘geeks’: 3}}, ‘geeks’: {‘geeks’: {‘para’: 7}}}
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}}, ‘para’: {‘geeks’: {‘geeks’: 3}}, ‘geeks’: {‘geeks’: {‘para’: 7}}}
Método n.º 2: uso de dictados predeterminados y enfoque recursivo
Python3
# 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)))
Salida:
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}}}
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}}}
Método #3: Usando reduce y getitem
Python3
# 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)))
Salida:
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}}}
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}}}
Publicación traducida automáticamente
Artículo escrito por garg_ak0109 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA