Python - Znajdź duplikaty na liście słowników i grupy nich

Question

nie jestem programistą, a także nowy w Pythonie, mam listę dicts pochodzących z pliku json:

# JSON file (film.json) 
[{"year": ["1999"], "director": ["Wachowski"], "film": ["The Matrix"], "price": ["19,00"]}, 
{"year": ["1994"], "director": ["Tarantino"], "film": ["Pulp Fiction"], "price": ["20,00"]}, 
{"year": ["2003"], "director": ["Tarantino"], "film": ["Kill Bill vol.1"], "price": ["10,00"]}, 
{"year": ["2003"], "director": ["Wachowski"], "film": ["The Matrix Reloaded"], "price": ["9,99"]}, 
{"year": ["1994"], "director": ["Tarantino"], "film": ["Pulp Fyction"], "price": ["15,00"]}, 
{"year": ["1994"], "director": ["E. de Souza"], "film": ["Street Fighter"], "price": ["2,00"]}, 
{"year": ["1999"], "director": ["Wachowski"], "film": ["The Matrix"], "price": ["20,00"]}, 
{"year": ["1982"], "director": ["Ridley Scott"], "film": ["Blade Runner"], "price": ["19,99"]}] 

mogę zaimportować json złożyć:

import json 
json_file = open('film.json') 
f = json.load(json_file) 

ale po tym nie jestem w stanie znaleźć wystąpienia w f i rozdzielić je w grupach tytuł filmu. To właśnie szukam osiągnąć:

## result grouped by 'film' 
#group 1 
{"year": ["1999"], "director": ["Wachowski"], "film": ["The Matrix"], "price": ["19,00"]} 
{"year": ["1999"], "director": ["Wachowski"], "film": ["The Matrix"], "price": ["20,00"]} 
#group 2 
{"year": ["1994"], "director": ["Tarantino"], "film": ["Pulp Fiction"], "price": ["20,00"]} 
{"year": ["1994"], "director": ["Tarantino"], "film": ["Pulp Fyction"], "price": ["15,00"]} 
#group X 
... 

albo lepiej:

new_dict = { 'group1':[[],[],...] , 'group2':[[],[],...] , 'groupX':[...] } 

W tej chwili jestem testowanie z zagnieżdżonego for ale bez powodzenia ..

Dziękuję .

uwaga: „miąższ fyction” jest poszukiwany błąd dla przyszłej realizacji z rozmytych dopasowania ciąg, bo teraz muszę tylko duplikaty Grouper „”

Uwaga2: z Pythona 2.x

Answer 1

Ponieważ dane nie jest posortowana, użyć collections.defaultdict() object materializować listę nowych kluczy, następnie kluczowych tytuł filmu:

from collections import defaultdict 

grouped = defaultdict(list) 

for film in f: 
    grouped[film['film'][0]].append(film) 

wartość film['film'][0] służy do grupy filmów. Musiałbyś stworzyć kanoniczną wersję tego klucza, gdybyś chciał użyć bardziej wyrafinowanego grupowania tytułów.

Demo:

>>> from collections import defaultdict 
>>> import json 
>>> with open('film.json') as film_file: 
...  f = json.load(film_file) 
... 
>>> grouped = defaultdict(list) 
>>> for film in f: 
...  grouped[film['film'][0]].append(film) 
... 
>>> grouped 
defaultdict(<type 'list'>, {u'Street Fighter': [{u'director': [u'E. de Souza'], u'price': [u'2,00'], u'film': [u'Street Fighter'], u'year': [u'1994']}], u'Pulp Fiction': [{u'director': [u'Tarantino'], u'price': [u'20,00'], u'film': [u'Pulp Fiction'], u'year': [u'1994']}], u'Pulp Fyction': [{u'director': [u'Tarantino'], u'price': [u'15,00'], u'film': [u'Pulp Fyction'], u'year': [u'1994']}], u'The Matrix': [{u'director': [u'Wachowski'], u'price': [u'19,00'], u'film': [u'The Matrix'], u'year': [u'1999']}, {u'director': [u'Wachowski'], u'price': [u'20,00'], u'film': [u'The Matrix'], u'year': [u'1999']}], u'Blade Runner': [{u'director': [u'Ridley Scott'], u'price': [u'19,99'], u'film': [u'Blade Runner'], u'year': [u'1982']}], u'Kill Bill vol.1': [{u'director': [u'Tarantino'], u'price': [u'10,00'], u'film': [u'Kill Bill vol.1'], u'year': [u'2003']}], u'The Matrix Reloaded': [{u'director': [u'Wachowski'], u'price': [u'9,99'], u'film': [u'The Matrix Reloaded'], u'year': [u'2003']}]}) 
>>> from pprint import pprint 
>>> pprint(dict(grouped)) 
{u'Blade Runner': [{u'director': [u'Ridley Scott'], 
        u'film': [u'Blade Runner'], 
        u'price': [u'19,99'], 
        u'year': [u'1982']}], 
u'Kill Bill vol.1': [{u'director': [u'Tarantino'], 
         u'film': [u'Kill Bill vol.1'], 
         u'price': [u'10,00'], 
         u'year': [u'2003']}], 
u'Pulp Fiction': [{u'director': [u'Tarantino'], 
        u'film': [u'Pulp Fiction'], 
        u'price': [u'20,00'], 
        u'year': [u'1994']}], 
u'Pulp Fyction': [{u'director': [u'Tarantino'], 
        u'film': [u'Pulp Fyction'], 
        u'price': [u'15,00'], 
        u'year': [u'1994']}], 
u'Street Fighter': [{u'director': [u'E. de Souza'], 
         u'film': [u'Street Fighter'], 
         u'price': [u'2,00'], 
         u'year': [u'1994']}], 
u'The Matrix': [{u'director': [u'Wachowski'], 
        u'film': [u'The Matrix'], 
        u'price': [u'19,00'], 
        u'year': [u'1999']}, 
       {u'director': [u'Wachowski'], 
        u'film': [u'The Matrix'], 
        u'price': [u'20,00'], 
        u'year': [u'1999']}], 
u'The Matrix Reloaded': [{u'director': [u'Wachowski'], 
          u'film': [u'The Matrix Reloaded'], 
          u'price': [u'9,99'], 
          u'year': [u'2003']}]} 

Stosując SoundEx filmom grupy byłoby proste:

from itertools import groupby, islice, ifilter 

_codes = ('bfpv', 'cgjkqsxz', 'dt', 'l', 'mn', 'r') 
_sounds = {c: str(i) for i, code in enumerate(_codes, 1) for c in code} 
_sounds.update(dict.fromkeys('aeiouy')) 
def soundex(word, _sounds=_sounds): 
    grouped = groupby(_sounds[c] for c in word.lower() if c in _sounds) 
    if _sounds.get(word[0].lower()): 
     next(grouped) # remove first group. 
    sdx = ''.join([k for k, g in islice((g for g in grouped if g[0]), 3)]) 
    return word[0].upper() + format(sdx, '<03') 

grouped_by_soundex = defaultdict(list) 
for film in f: 
    grouped_by_soundex[soundex(film['film'][0])].append(film) 

powodując:

>>> pprint(dict(grouped_by_soundex)) 
{u'B436': [{u'director': [u'Ridley Scott'], 
      u'film': [u'Blade Runner'], 
      u'price': [u'19,99'], 
      u'year': [u'1982']}], 
u'K414': [{u'director': [u'Tarantino'], 
      u'film': [u'Kill Bill vol.1'], 
      u'price': [u'10,00'], 
      u'year': [u'2003']}], 
u'P412': [{u'director': [u'Tarantino'], 
      u'film': [u'Pulp Fiction'], 
      u'price': [u'20,00'], 
      u'year': [u'1994']}, 
      {u'director': [u'Tarantino'], 
      u'film': [u'Pulp Fyction'], 
      u'price': [u'15,00'], 
      u'year': [u'1994']}], 
u'S363': [{u'director': [u'E. de Souza'], 
      u'film': [u'Street Fighter'], 
      u'price': [u'2,00'], 
      u'year': [u'1994']}], 
u'T536': [{u'director': [u'Wachowski'], 
      u'film': [u'The Matrix'], 
      u'price': [u'19,00'], 
      u'year': [u'1999']}, 
      {u'director': [u'Wachowski'], 
      u'film': [u'The Matrix Reloaded'], 
      u'price': [u'9,99'], 
      u'year': [u'2003']}, 
      {u'director': [u'Wachowski'], 
      u'film': [u'The Matrix'], 
      u'price': [u'20,00'], 
      u'year': [u'1999']}]} 

Answer 2

Jeśli to jednorazowe i ja w pośpiechu zrobię to w ten sposób. Zakładając, że na tym przykładzie, że lista słowników jest lod, i że tytuł filmu będzie tylko kiedykolwiek być lista z jednej pozycji

new_dict = {k:[d for d in lod if d.get('film')[0] == k] for k in set(d.get('film')[0] for d in l)} 

Aby uczynić go bardziej czytelnym i wyjaśnić, co robi, to samo uszkodzony się, ponownie lista słowników jest:

#get all the unique film names 
# note: the [0] is because its a list for the title, and set doesn't work with lists, 
#so we're just taking the first one for this example. 
films = set(d.get('film')[0] for d in lod) 


#create a dictionary 
new_dict = {} 

#iterate over the unique film names 
for k in films: 
    #make a list of all the films that match the name we're on 
    filmswiththisname = [d for d in lod if d.get('film')[0] == k] 
    #add the list of films to the new dictionary with the film name as the key. 
    new_dict[k] = filmswiththisname