comprehension - python dictionary methods

Python-Aplanar la lista de diccionarios

Question

Lista de diccionarios:

data = [{
         ''a'':{''l'':''Apple'',
                ''b'':''Milk'',
                ''d'':''Meatball''},
         ''b'':{''favourite'':''coke'',
              ''dislike'':''juice''}
         },
         {
         ''a'':{''l'':''Apple1'',
                ''b'':''Milk1'',
                ''d'':''Meatball2''},
         ''b'':{''favourite'':''coke2'',
              ''dislike'':''juice3''}
         }, ...
]

Necesito unirme a todos los diccionarios anidados para alcanzar el resultado esperado:

 [{''d'': ''Meatball'', ''b'': ''Milk'', ''l'': ''Apple'', ''dislike'': ''juice'', ''favourite'': ''coke''},
  {''d'': ''Meatball2'', ''b'': ''Milk1'', ''l'': ''Apple1'', ''dislike'': ''juice3'', ''favourite'': ''coke2''}]

Intento la comprensión de la lista anidada, pero no puedo unir a dict juntos:

L = [y for x in data for y in x.values()]
print (L)

[{''d'': ''Meatball'', ''b'': ''Milk'', ''l'': ''Apple''}, 
 {''dislike'': ''juice'', ''favourite'': ''coke''}, 
{''d'': ''Meatball2'', ''b'': ''Milk1'', ''l'': ''Apple1''}, 
 {''dislike'': ''juice3'', ''favourite'': ''coke2''}]

Estoy buscando la solución más rápida.

Answer 1

Puede hacer esto con 2 bucles anidados, y dict.update() para agregar diccionarios internos a un diccionario temporal y agregarlo al final:

L = []
for d in data:
    temp = {}
    for key in d:
        temp.update(d[key])

    L.append(temp)

# timeit ~1.4
print(L)

Qué salidas:

[{''l'': ''Apple'', ''b'': ''Milk'', ''d'': ''Meatball'', ''favourite'': ''coke'', ''dislike'': ''juice''}, {''l'': ''Apple1'', ''b'': ''Milk1'', ''d'': ''Meatball2'', ''favourite'': ''coke2'', ''dislike'': ''juice3''}]

Answer 2

Puede usar functools.reduce junto con una lista de comprensión simple para aplanar la lista de dictos

>>> from functools import reduce 

>>> data = [{''b'': {''dislike'': ''juice'', ''favourite'': ''coke''}, ''a'': {''l'': ''Apple'', ''b'': ''Milk'', ''d'': ''Meatball''}}, {''b'': {''dislike'': ''juice3'', ''favourite'': ''coke2''}, ''a'': {''l'': ''Apple1'', ''b'': ''Milk1'', ''d'': ''Meatball2''}}]
>>> [reduce(lambda x,y: {**x,**y},d.values()) for d in data]
>>> [{''dislike'': ''juice'', ''l'': ''Apple'', ''d'': ''Meatball'', ''b'': ''Milk'', ''favourite'': ''coke''}, {''dislike'': ''juice3'', ''l'': ''Apple1'', ''d'': ''Meatball2'', ''b'': ''Milk1'', ''favourite'': ''coke2''}]

El tiempo de referencia es el siguiente:

>>> import timeit
>>> setup = """
      from functools import reduce
      data = [{''b'': {''dislike'': ''juice'', ''favourite'': ''coke''}, ''a'': {''l'': ''Apple'', ''b'': ''Milk'', ''d'': ''Meatball''}}, {''b'': {''dislike'': ''juice3'', ''favourite'': ''coke2''}, ''a'': {''l'': ''Apple1'', ''b'': ''Milk1'', ''d'': ''Meatball2''}}]
  """
>>> min(timeit.Timer("[reduce(lambda x,y: {**x,**y},d.values()) for d in data]",setup=setup).repeat(3,1000000))
>>> 1.525032774952706

Tiempo de referencia de otras respuestas en mi máquina

>>> setup = """
        data = [{''b'': {''dislike'': ''juice'', ''favourite'': ''coke''}, ''a'': {''l'': ''Apple'', ''b'': ''Milk'', ''d'': ''Meatball''}}, {''b'': {''dislike'': ''juice3'', ''favourite'': ''coke2''}, ''a'': {''l'': ''Apple1'', ''b'': ''Milk1'', ''d'': ''Meatball2''}}]
    """
>>> min(timeit.Timer("[{k: v for x in d.values() for k, v in x.items()} for d in data]",setup=setup).repeat(3,1000000))
>>> 2.2488374650129117

>>> min(timeit.Timer("[{k: x[k] for x in d.values() for k in x} for d in data]",setup=setup).repeat(3,1000000))
>>> 1.8990078769857064

>>> code = """
      L = []
      for d in data:
          temp = {}
          for key in d:
              temp.update(d[key])

          L.append(temp)
    """

>>> min(timeit.Timer(code,setup=setup).repeat(3,1000000))
>>> 1.4258553800173104

>>> setup = """
      from itertools import chain
      data = [{''b'': {''dislike'': ''juice'', ''favourite'': ''coke''}, ''a'': {''l'': ''Apple'', ''b'': ''Milk'', ''d'': ''Meatball''}}, {''b'': {''dislike'': ''juice3'', ''favourite'': ''coke2''}, ''a'': {''l'': ''Apple1'', ''b'': ''Milk1'', ''d'': ''Meatball2''}}]
    """
>>> min(timeit.Timer("[dict(chain(*map(dict.items, d.values()))) for d in data]",setup=setup).repeat(3,1000000))
>>> 3.774383604992181

Answer 3

Puedes hacer lo siguiente, usando itertools.chain :

>>> from itertools import chain
# timeit: ~3.40
>>> [dict(chain(*map(dict.items, d.values()))) for d in data]
[{''l'': ''Apple'', 
  ''b'': ''Milk'', 
  ''d'': ''Meatball'', 
  ''favourite'': ''coke'', 
  ''dislike'': ''juice''}, 
 {''l'': ''Apple1'', 
  ''b'': ''Milk1'', 
  ''dislike'': ''juice3'', 
  ''favourite'': ''coke2'', 
  ''d'': ''Meatball2''}]

El uso de chain , map , * hace que esta expresión sea una abreviatura de la siguiente comprensión anidada que realmente funciona mejor en mi sistema (Python 3.5.2) y no es mucho más larga:

# timeit: ~2.04
[{k: v for x in d.values() for k, v in x.items()} for d in data]
# Or, not using items, but lookup by key
# timeit: ~1.67
[{k: x[k] for x in d.values() for k in x} for d in data]

Nota:

RoadRunner''s enfoque RoadRunner''s bucle y actualización de RoadRunner''s supera a estas dos líneas en el timeit: ~1.37

Answer 4

Si tiene diccionarios anidados con solo las teclas ''a'' y ''b'', sugiero la siguiente solución que encuentro rápida y muy fácil de entender (para facilitar la lectura):

L = [x[''a''] for x in data]
b = [x[''b''] for x in data]

for i in range(len(L)):
    L[i].update(b[i])

# timeit ~1.4

print(L)