我不是程序员,也是 python 的新手,我有一个来自 json 文件的字典列表:
# 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"]}]
我可以导入 json 文件:
import json
json_file = open('film.json')
f = json.load(json_file)
但在那之后我无法在 f
中找到事件并按电影标题将它们分组。
这就是我要实现的目标:
## 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
...
或者更好:
new_dict = { 'group1':[[],[],...] , 'group2':[[],[],...] , 'groupX':[...] }
目前我正在使用嵌套的 for
进行测试,但没有运气..
谢谢。
注意:“pulp fyction”是 future 使用模糊字符串匹配实现时想要犯的错误,现在我只需要一个“duplicates grouper”
注2:使用python 2.x
最佳答案
因为你的数据没有排序,所以使用 collections.defaultdict()
object实现新键列表,然后按电影标题键:
from collections import defaultdict
grouped = defaultdict(list)
for film in f:
grouped[film['film'][0]].append(film)
film['film'][0]
值用于对电影进行分组。如果您想使用更复杂的标题分组,则必须创建该 key 的规范版本。
演示:
>>> 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']}]}
使用 SoundEx对电影进行分组非常简单:
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)
导致:
>>> 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']}]}
关于Python - 在字典列表中查找重复项并将它们分组,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19011124/