我需要創(chuàng)建一個字典，其鍵是來自先前創(chuàng)建的字典的值。我以前的字典是：{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])}這個字典的形式是 {key: (id, string, tags, likes, dislikes)} 我需要創(chuàng)建一個字典，它的鍵是第一個字典中的標(biāo)簽，它的值是包含標(biāo)簽，以字典的形式呈現(xiàn)。例如，如果我們使用 tag 'SnowMan'，新字典應(yīng)該是這樣的：{'SnowMan': {400: ['Does not want to build a %SnowMan %StopAsking', 'LOLZ BELLE. %StockholmeSyndrome %SnowMan']}}或者，如果我們使用 tag 'StockholmeSyndrome'，新字典應(yīng)該是這樣的：{'StockholmeSyndrome': { 500: ["Actually nm. This isn't so bad lolz :P %StockholmeSyndrome"], 400: ['LOLZ BELLE. %StockholmeSyndrome %SnowMan']}}最后，新字典需要將之前字典中的所有標(biāo)簽都包含為鍵，包括沒有標(biāo)簽的事件。我真的被困在這個問題上，有人可以提供一些指導(dǎo)嗎？
3 回答
冉冉說
TA貢獻(xiàn)1877條經(jīng)驗 獲得超1個贊
使用collections.defualtdict您可以迭代并附加到嵌套字典結(jié)構(gòu)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 的子類dict，因此您通常不需要進(jìn)一步操作。但是，如果您需要常規(guī)dict對象，則可以使用遞歸函數(shù)：
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']}}
繁星淼淼
TA貢獻(xiàn)1775條經(jīng)驗 獲得超11個贊
那這個呢 ？
keys = ['name', 'last_name', 'phone_number', 'email']
dict2 = {x:dict1[x] for x in keys}
海綿寶寶撒
TA貢獻(xiàn)1809條經(jīng)驗 獲得超8個贊
在查看了您的要求后，我已使用以下代碼更新了我之前的帖子。我相信這就是您正在努力實(shí)現(xiàn)的目標(biāo)。唯一需要的導(dǎo)入是來自 collections 模塊的OrderedDict。
from collections import OrderedDict
from pprint import PrettyPrinter as pp
str_tag_dict = {}
for v in d.values():
str_tag_dict[v[1]] = v[2]
def same_tags_and_ids(str1, str2):
dd = {str1:str_tag_dict[str1],
str2:str_tag_dict[str2]}
tag = []
v = dd[str1]
for vv in v:
for ks in dd.keys():
if vv in dd[ks]:
tag.append(vv)
for tg in tag:
if tag.count(tg) > 1:
return tg
return None
ordp = OrderedDict(sorted(d.items()))
def same_tags_diff_ids():
x = []
y = []
my_str = []
for v in ordp.values():
x.append(v[2][0])
x.append(v[0])
x.append(v[1])
for i in x:
if x.count(i) > 1 and type(i) == str:
y.append(i)
for i in range(len(x)):
if x[i] == list(set(y))[0]:
my_str.append({list(set(y))[0]:{x[i + 1]:[x[i + 2]]}})
d={}
k1, v1 = list(my_str[0][list(my_str[0])[0]].items())[0]
k2, v2 = list(my_str[1][list(my_str[1])[0]].items())[0]
d.update({list(my_str[0].keys())[0]:{k1:v1, k2:v2}})
if not d == None:
return d
else:
return None
dup_tag_str = same_tags_diff_ids()
for k, v in ordp.items():
del d[k]
for tag in v[2]:
d[tag] = {v[0]:[v[1]]}
ordp = OrderedDict(sorted(d.items()))
ids= [list(v.keys())[0] for v in ordp.values()]
strs= [list(v.values())[0][0] for v in ordp.values()]
x = 0
for n in range(len(ordp)):
tmp_keys = []
for k, v in ordp.items():
if not strs[x] == list(v.values())[0][0]:
if ids[x] == list(v.keys())[0]:
key = same_tags_and_ids(strs[x], list(v.values())[0][0])
if not key == None:
tmp_keys.append(key)
if tmp_keys.count(key) == 1:
continue
else:
ordp[key][list(v.keys())[0]].append(strs[x])
x += 1
dx = dup_tag_str
dy = dict(ordp)
dy.update(dx)
pp().pprint(dy)
輸出：
{'': {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: ['LOLZ BELLE. %StockholmeSyndrome %SnowMan',
'Does not want to build a %SnowMan %StopAsking']},
'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']}}
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