概述

# JSON file (film.json)
[{"year": ["1999"],"director": ["Wachowski"],"film": ["The Matrix"],"price": ["19,00"]},{"year": ["1994"],"director": ["Tarantino"],"film": ["Pulp Fiction"],"price": ["20,{"year": ["2003"],"film": ["Kill Bill vol.1"],"price": ["10,"film": ["The Matrix Reloaded"],"price": ["9,99"]},"film": ["Pulp Fyction"],"price": ["15,"director": ["E. de Souza"],"film": ["Street fighter"],"price": ["2,{"year": ["1999"],{"year": ["1982"],"director": ["Ridley Scott"],"film": ["Blade Runner"],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,并按电影片名分组.
这就是我想要实现的目标：

## result grouped by 'film'
#group 1
{"year": ["1999"],00"]}
{"year": ["1999"],00"]}
#group 2
{"year": ["1994"],00"]}
{"year": ["1994"],00"]}
#group X
 ...

或更好：

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

目前我正在测试嵌套,但没有运气..

谢谢.

注意：“纸浆fyction”是未来实现的模糊字符串匹配的错误,现在我只需要一个’重复的石斑鱼’

note2：使用python 2.x.

from collections import defaultdict

grouped = defaultdict(list)

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

电影[‘电影’] [0]值用于分组电影.如果您想使用更复杂的标题分组,则必须创建该密钥的规范版本.

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,u'film': [u'Pulp Fiction'],u'Pulp Fyction': [{u'director': [u'Tarantino'],u'price': [u'15,u'film': [u'Pulp Fyction'],u'The Matrix': [{u'director': [u'Wachowski'],u'price': [u'19,u'film': [u'The Matrix'],u'year': [u'1999']},{u'director': [u'Wachowski'],u'year': [u'1999']}],u'Blade Runner': [{u'director': [u'Ridley Scott'],99'],u'film': [u'Blade Runner'],u'year': [u'1982']}],u'Kill Bill vol.1': [{u'director': [u'Tarantino'],u'price': [u'10,u'film': [u'Kill Bill vol.1'],u'year': [u'2003']}],u'The Matrix Reloaded': [{u'director': [u'Wachowski'],u'price': [u'9,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'Street fighter': [{u'director': [u'E. de Souza'],u'year': [u'2003']}]}

使用SoundEx分组电影将如下：

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'K414': [{u'director': [u'Tarantino'],u'P412': [{u'director': [u'Tarantino'],u'year': [u'1994']},{u'director': [u'Tarantino'],u'S363': [{u'director': [u'E. de Souza'],u'T536': [{u'director': [u'Wachowski'],u'year': [u'2003']},u'year': [u'1999']}]}

总结

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