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how to take all tweets in a hashtag with tweepy?

I'm trying to take every open tweets in a hashtag but my code does not go further than 299 tweets.

I also trying to take tweets from a specific time line like tweets only in May 2015 and July 2016. Are there any way to do it in the main process or should I write a little code for it?

Here is my code:

# 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
like image 647
shuetisha.dev Avatar asked Dec 24 '22 16:12

shuetisha.dev


1 Answers

Have a look at this: https://tweepy.readthedocs.io/en/v3.5.0/cursor_tutorial.html

And try this:

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

In your case you have a max number of tweets to get, so as per the linked tutorial you could do:

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

If you want tweets after a given ID, you can also pass that argument.

like image 168
Paco H. Avatar answered Jan 14 '23 14:01

Paco H.