python - 如何使用以前字典中的键和值创建字典？

Question

我需要创建一个字典，其键值来自先前创建的字典。我以前的词典是:

{100000: (400, 'Does not want to build a %SnowMan %StopAsking', ['SnowMan', 'StopAsking'], [100, 200, 300], [400, 500]), 
100001: (200, 'Make the ocean great again.', [''], [], [400]), 
100002: (500, "Help I'm being held captive by a beast!  %OhNoes", ['OhNoes'], [400], [100, 200, 300]), 
100003: (500, "Actually nm. This isn't so bad lolz :P %StockholmeSyndrome", ['StockholmeSyndrome'], [400, 100], []), 
100004: (300, 'If some random dude offers to %ShowYouTheWorld do yourself a favour and %JustSayNo.', ['ShowYouTheWorld', 'JustSayNo'], [500, 200], [400]), 
100005: (400, 'LOLZ BELLE.  %StockholmeSyndrome  %SnowMan', ['StockholmeSyndrome', 'SnowMan'], [], [200, 300, 100, 500])}

这个字典的形式是{key: (id, string, tags, likes, dislikes)}我需要创建的是一个字典，其键是第一个字典中的标签，其值是包含标签的 id 和字符串，以字典的形式呈现。

例如，如果我们采用标签'SnowMan'，新字典应该如下所示:

{'SnowMan': {400: ['Does not want to build a %SnowMan %StopAsking', 'LOLZ BELLE.  %StockholmeSyndrome  %SnowMan']}}

或者，如果我们使用标签 'StockholmeSyndrome'，新字典应该如下所示:

{'StockholmeSyndrome': {
        500: ["Actually nm. This isn't so bad lolz :P %StockholmeSyndrome"], 
        400: ['LOLZ BELLE.  %StockholmeSyndrome  %SnowMan']}}

最后，新字典需要包含之前字典中的所有标签作为键，包括没有标签的事件。

我真的被这个问题困住了，谁能提供一些指导？

Answer 1

使用 collections.defualtdict您可以迭代并附加到嵌套字典结构中的列表:

from collections import defaultdict

dd = defaultdict(lambda: defaultdict(list))
for id_, text, tags, likes, dislikes in d.values():
    for tag in tags:
        dd[tag][id_].append(text)

print(dd)

defaultdict(<function __main__.<lambda>>,
            {'': defaultdict(list, {200: ['Make the ocean great again.']}),
             'JustSayNo': defaultdict(list,
                         {300: ['If some random dude offers to %ShowYouTheWorld do yourself a favour and %JustSayNo.']}),
             'OhNoes': defaultdict(list,
                         {500: ["Help I'm being held captive by a beast!  %OhNoes"]}),
             'ShowYouTheWorld': defaultdict(list,
                         {300: ['If some random dude offers to %ShowYouTheWorld do yourself a favour and %JustSayNo.']}),
             'SnowMan': defaultdict(list,
                         {400: ['Does not want to build a %SnowMan %StopAsking',
                           'LOLZ BELLE.  %StockholmeSyndrome  %SnowMan']}),
             'StockholmeSyndrome': defaultdict(list,
                         {400: ['LOLZ BELLE.  %StockholmeSyndrome  %SnowMan'],
                          500: ["Actually nm. This isn't so bad lolz :P %StockholmeSyndrome"]}),
             'StopAsking': defaultdict(list,
                         {400: ['Does not want to build a %SnowMan %StopAsking']})})

defaultdict 是 dict 的子类，因此您通常不需要进一步操作。但是，如果您需要一个常规的 dict 对象，您可以使用递归函数:

def default_to_regular_dict(d):
    """Convert nested defaultdict to regular dict of dicts."""
    if isinstance(d, defaultdict):
        d = {k: default_to_regular_dict(v) for k, v in d.items()}
    return d

res = default_to_regular_dict(dd)

print(res)

{'': {200: ['Make the ocean great again.']},
 'JustSayNo': {300: ['If some random dude offers to %ShowYouTheWorld do yourself a favour and %JustSayNo.']},
 'OhNoes': {500: ["Help I'm being held captive by a beast!  %OhNoes"]},
 'ShowYouTheWorld': {300: ['If some random dude offers to %ShowYouTheWorld do yourself a favour and %JustSayNo.']},
 'SnowMan': {400: ['Does not want to build a %SnowMan %StopAsking',
   'LOLZ BELLE.  %StockholmeSyndrome  %SnowMan']},
 'StockholmeSyndrome': {400: ['LOLZ BELLE.  %StockholmeSyndrome  %SnowMan'],
  500: ["Actually nm. This isn't so bad lolz :P %StockholmeSyndrome"]},
 'StopAsking': {400: ['Does not want to build a %SnowMan %StopAsking']}}