根据字典中的键和值修改循环

Question

我是 python 的新手，我写了下面的代码，在字典中搜索，做一些事情，清除字典中的旧项目并用新的键和值更新字典，并在没有注意添加到字典时中断（它是空的），怎么能我修改我的代码来做这个过程？

#since_id - Returns results with an ID greater than 
#(that is, more recent than) the specified ID. There are limits to the 
#number of Tweets which can be accessed through the API.
# If the limit of Tweets has occured since the since_id,
# the since_id will be forced to the oldest ID available. 
# max_id - Returns results with an ID less than (that is, older than) 
#or equal to the specified ID.

Dict2 = dict({'@TweeetLorraine':1392217841680764931})
d2 = {}
rep=[] 
from tqdm import tqdm
for key, value in tqdm(Dict2.items()):
  for i in tweepy.Cursor(api.search,
                     q='to:{} -filter:retweets"'.format(key),lang="en"
                     ,since_id=value,tweet_mode='extended',
                     wait_on_rate_limit=True,
                     wait_on_rate_limit_notify=True).items(50):
                     if (i.in_reply_to_status_id == value):
                       rep.append(i)


                       from pandas.io.json import json_normalize
                       dfflat = pd.DataFrame()
                       for tweet in rep:
                         df_for_tweet = json_normalize(tweet._json)
                         dfflat=dfflat.append(df_for_tweet,ignore_index=True,sort=True)

                         d2.update(zip(dfflat["user.screen_name"].tolist(), dfflat["id"].tolist()))





d2 ```

Answer 1

您可以为此使用while循环：

#since_id - Returns results with an ID greater than 
#(that is, more recent than) the specified ID. There are limits to the 
#number of Tweets which can be accessed through the API.
# If the limit of Tweets has occured since the since_id,
# the since_id will be forced to the oldest ID available. 
# max_id - Returns results with an ID less than (that is, older than) 
#or equal to the specified ID.

Dict2 = dict({'@TweeetLorraine':1392217841680764931})
d2 = {}
rep=[] 
from tqdm import tqdm
for key, value in tqdm(Dict2.items()):
  for i in tweepy.Cursor(api.search,
                     q='to:{} -filter:retweets"'.format(key),lang="en"
                     ,since_id=value,tweet_mode='extended',
                     wait_on_rate_limit=True,
                     wait_on_rate_limit_notify=True).items(50):
                     if (i.in_reply_to_status_id == value):
                       rep.append(i)


                       from pandas.io.json import json_normalize
                       dfflat = pd.DataFrame()
                       for tweet in rep:
                         df_for_tweet = json_normalize(tweet._json)
                         dfflat=dfflat.append(df_for_tweet,ignore_index=True,sort=True)

                         d2.update(zip(dfflat["user.screen_name"].tolist(), dfflat["id"].tolist()))





d2

对于您的用例，这里大致是执行您描述的代码，使用 map 有更好的方法，如果您想了解更多信息，我让您搜索它。

另外，我不确定您是要完全清除 dict 还是只清除当前的“i”，但我认为您可以根据自己的实际需要修改以下代码段

mydict = initial_dict
# while there is something in the dictionary
while mydict:
    value_searched = None
    for key, value in mydict.items():
        for i in tweepy.Cursor(api.search,
                     q='to:{} -filter:retweets"'.format(key),lang="en"
                     ,since_id=value,tweet_mode='extended',
                     wait_on_rate_limit=True,
                     wait_on_rate_limit_notify=True).items(50):
                     if (i.in_reply_to_status_id == value):
                       replies3.append(i)
                       value_searched = i
                       break
         break

    # create new dict from value retrieved
    mydict = {"@" +value_searched.user.screen_name : value_searched.id_str} 

Edit2：使用递归

def tweepy_stub(key, value):
    if key == "TweeetLorraine" and value == 1392217841680764931:
        return [
            ("AlexBC997", 1392385334155956226),
            ("ChapinDolores", 1392432099945238529),
        ]
    elif key == "AlexBC997" and value == 1392385334155956226:
        return [("test", 139238533415595852)]
    elif ("ChapinDolores", 1392432099945238529):
        return []


def recursive(list_values, nb_recursion):
    mydict = {}
    if list_values == None or nb_recursion == 0:
        return mydict
    else:
        for name_user, tweet_id in list_values:
            mydict[(name_user, tweet_id)] = recursive(
                retrieve_direct_reply_stub(name_user, tweet_id), nb_recursion - 1
            )
        return mydict

class stub_tweepy_answer:
    def __init__(self, status_id) -> None:
        self.in_reply_to_status_id = status_id

def retrieve_direct_reply_stub(name_user, tweepy_id):
    rep = []
    d2 = []
    return tweepy_stub(name_user, tweepy_id)

def retrieve_direct_reply(name_user, tweet_id):
    rep = []
    d2 = []
    for i in tweepy_stub(name_user, tweet_id):
        val = i
        if (i.in_reply_to_status_id == tweet_id):
            rep.append(i)
            from pandas.io.json import json_normalize
            dfflat = pd.DataFrame()
            for tweet in rep:
                df_for_tweet = json_normalize(tweet._json)
                dfflat=dfflat.append(df_for_tweet,ignore_index=True,sort=True)

                d2.append(zip(dfflat["user.screen_name"].tolist(), dfflat["id"].tolist()))
    return d2

#print(retrieve_direct_reply_stub("TweeetLorraine", 1392217841680764931))

elem = [("TweeetLorraine", 1392217841680764931)]
print(recursive(elem, 3))