Python在字典列表中查找重复项并对其进行分组

Question

我不是程序员，也是python新手，我有一个来自json文件的dict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"]}]

我可以通过以下方式导入json文件：

^{pr2}$

但在那之后，我就无法找到f中的事件，并按片名分组。 这就是我想要达到的目标：

## 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
 ...

或更好：

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

目前我正在用嵌套的for进行测试，但是没有运气。。在

谢谢。在

注意：“pulpfyAction”是未来实现模糊字符串匹配时的一个错误，现在我只需要一个“重复的grouper”

注2：使用python2.x

Answer 1

由于您的数据未排序，请使用^{} object来具体化新键的列表，然后按胶片标题键：

from collections import defaultdict

grouped = defaultdict(list)

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

{cd2>值是用来分组的。如果要使用更复杂的标题分组，则必须创建该密钥的规范版本。在

演示：

^{pr2}$

使用SoundEx对胶片进行分组非常简单：

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']}]}

Answer 2

如果是一次性的，而且我很匆忙，我会这样做的。在这个例子中，假设您的字典列表是lod，并且电影标题永远只包含一个条目

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)}

为了使它更具可读性，并解释它在做什么，同样的事情发生了，同样的字典列表是lod:

^{pr2}$