python之如何使用 tweepy 获取标签中的所有推文

我试图将所有打开的推文都放入一个标签中，但我的代码不会超过 299 条推文。

我还尝试从特定时间线获取推文，例如仅在 2015 年 5 月和 2016 年 7 月的推文。是否有任何方法可以在主流程中执行此操作，或者我是否应该为其编写一些代码？

这是我的代码:

# if this is the first time, creates a new array which 
# will store max id of the tweets for each keyword 
if not os.path.isfile("max_ids.npy"): 
    max_ids = np.empty(len(keywords)) 
    # every value is initialized as -1 in order to start from the beginning the first time program run 
    max_ids.fill(-1) 
else: 
    max_ids = np.load("max_ids.npy")  # loads the previous max ids 
 
# if there is any new keywords added, extends the max_ids array in order to correspond every keyword 
if len(keywords) > len(max_ids): 
    new_indexes = np.empty(len(keywords) - len(max_ids)) 
    new_indexes.fill(-1) 
    max_ids = np.append(arr=max_ids, values=new_indexes) 
 
count = 0 
for i in range(len(keywords)): 
    since_date="2015-01-01" 
    sinceId = None 
    tweetCount = 0 
    maxTweets = 5000000000000000000000  # maximum tweets to find per keyword 
    tweetsPerQry = 100 
    searchQuery = "#{0}".format(keywords[i]) 
    while tweetCount < maxTweets: 
        if max_ids[i] < 0: 
                if (not sinceId): 
                    new_tweets = api.search(q=searchQuery, count=tweetsPerQry) 
                else: 
                    new_tweets = api.search(q=searchQuery, count=tweetsPerQry, 
                                            since_id=sinceId) 
        else: 
                if (not sinceId): 
                    new_tweets = api.search(q=searchQuery, count=tweetsPerQry, 
                                            max_id=str(max_ids - 1)) 
                else: 
                    new_tweets = api.search(q=searchQuery, count=tweetsPerQry, 
                                            max_id=str(max_ids - 1), 
                                            since_id=sinceId) 
        if not new_tweets: 
            print("Keyword: {0}      No more tweets found".format(searchQuery)) 
            break 
        for tweet in new_tweets: 
            count += 1 
            print(count) 
 
            file_write.write( 
                       . 
                       . 
                       . 
                         ) 
 
            item = { 
                . 
                . 
                . 
                . 
                . 
            } 
 
            # instead of using mongo's id for _id, using tweet's id 
            raw_data = tweet._json 
            raw_data["_id"] = tweet.id 
            raw_data.pop("id", None) 
 
            try: 
                db["Tweets"].insert_one(item) 
            except pymongo.errors.DuplicateKeyError as e: 
                print("Already exists in 'Tweets' collection.") 
            try: 
                db["RawTweets"].insert_one(raw_data) 
            except pymongo.errors.DuplicateKeyError as e: 
                print("Already exists in 'RawTweets' collection.") 
 
        tweetCount += len(new_tweets) 
        print("Downloaded {0} tweets".format(tweetCount)) 
        max_ids[i] = new_tweets[-1].id 
 
np.save(arr=max_ids, file="max_ids.npy")  # saving in order to continue mining from where left next time program run 

请您参考如下方法:

看看这个:https://tweepy.readthedocs.io/en/v3.5.0/cursor_tutorial.html

然后试试这个:

import tweepy 
 
auth = tweepy.OAuthHandler(CONSUMER_TOKEN, CONSUMER_SECRET) 
api = tweepy.API(auth) 
 
for tweet in tweepy.Cursor(api.search, q='#python', rpp=100).items(): 
    # Do something 
    pass 

在您的情况下，您可以获取最大数量的推文，因此根据链接的教程，您可以执行以下操作:

import tweepy 
 
MAX_TWEETS = 5000000000000000000000 
 
auth = tweepy.OAuthHandler(CONSUMER_TOKEN, CONSUMER_SECRET) 
api = tweepy.API(auth) 
 
for tweet in tweepy.Cursor(api.search, q='#python', rpp=100).items(MAX_TWEETS): 
    # Do something 
    pass 

如果您想要在给定 ID 之后发送推文，您也可以传递该参数。