[英]How to get the replies for a given tweet with tweepy and python?
After some searching I found a way to download the recent tweets from a given user. 经过一些搜索,我找到了一种从给定用户下载最新推文的方法。 Now I want to get the replies for each tweet.
现在,我想获取每个推文的回复。 I know that Twitter API does not provide an endpoint to get replies of a tweet unless we have a premium account.
我知道,除非我们有高级帐户,否则Twitter API不会提供端点来获取推文的回复。 But I could find some workarounds in the internet.
但是我可以在互联网上找到一些解决方法。 I found a way to get few tweets and their replies using Getting tweet replies to a particular tweet from a particular user .
我找到了一种使用获取特定用户对特定tweet的回复的方法来获取少量tweet及其回复的方法。 This code is also given below.
此代码也在下面给出。
How can I modify my code (getData.py) to save the replies of each tweet along with the tweet in the csv? 如何修改我的代码(getData.py),以将每个推文的回复与推文一起保存在csv中?
My code to download a user's tweets as csv (getData.py)
我的将用户鸣叫下载为csv的代码(getData.py)
import tweepy
import csv
# Twitter API credentials
consumer_key = "###########"
consumer_secret = "################"
access_key = "#################"
access_secret = "#####################"
def get_all_tweets(screen_name):
# Twitter only allows access to a users most recent 3240 tweets with this method
# authorize twitter, initialize tweepy
auth = tweepy.OAuthHandler(consumer_key, consumer_secret)
auth.set_access_token(access_key, access_secret)
api = tweepy.API(auth)
# initialize a list to hold all the tweepy Tweets
alltweets = []
# make initial request for most recent tweets (200 is the maximum allowed count)
new_tweets = api.user_timeline(screen_name=screen_name, count=200)
# save most recent tweets
alltweets.extend(new_tweets)
# save the id of the oldest tweet less one
oldest = alltweets[-1].id - 1
# keep grabbing tweets until there are no tweets left to grab
while len(new_tweets) > 0:
print
"getting tweets before %s" % (oldest)
# all subsiquent requests use the max_id param to prevent duplicates
new_tweets = api.user_timeline(screen_name=screen_name, count=200, max_id=oldest)
# save most recent tweets
alltweets.extend(new_tweets)
# update the id of the oldest tweet less one
oldest = alltweets[-1].id - 1
print
"...%s tweets downloaded so far" % (len(alltweets))
# transform the tweepy tweets into a 2D array that will populate the csv
outtweets = [[tweet.id_str, tweet.created_at, tweet.text.encode("utf-8"), tweet.favorite_count, tweet.retweet_count]
for tweet in alltweets]
# write the csv
with open('%s_tweets.csv' % screen_name, mode='w', encoding='utf-8') as f:
writer = csv.writer(f)
writer.writerow(["id", "created_at", "text"])
writer.writerows(outtweets)
pass
def main():
get_all_tweets("tartecosmetics")
if __name__ == '__main__':
main()
How I get the replies for a given tweet
如何获得给定推文的回复
This code will fetch 10 recent tweets of an user(name) along with the replies to that particular tweet. 该代码将获取用户(名称)最近的10条推文以及对该特定推文的答复。
replies=[]
non_bmp_map = dict.fromkeys(range(0x10000, sys.maxunicode + 1), 0xfffd)
for full_tweets in tweepy.Cursor(api.user_timeline,screen_name='tartecosmetics',timeout=999999).items(10):
for tweet in tweepy.Cursor(api.search,q='to:'+'tartecosmetics',result_type='recent',timeout=999999).items(1000):
if hasattr(tweet, 'in_reply_to_status_id_str'):
if (tweet.in_reply_to_status_id_str==full_tweets.id_str):
replies.append(tweet.text)
print("Tweet :",full_tweets.text.translate(non_bmp_map))
for elements in replies:
print("Replies :",elements)
replies.clear()
user_name = "@nameofuser"
replies = tweepy.Cursor(api.search, q='to:{}'.format(user_name),
since_id=tweet_id, tweet_mode='extended').items()
while True:
try:
reply = replies.next()
if not hasattr(reply, 'in_reply_to_status_id_str'):
continue
if reply.in_reply_to_status_id == tweet_id:
logging.info("reply of tweet:{}".format(reply.full_text))
except tweepy.RateLimitError as e:
logging.error("Twitter api rate limit reached".format(e))
time.sleep(60)
continue
except tweepy.TweepError as e:
logging.error("Tweepy error occured:{}".format(e))
break
except StopIteration:
break
except Exception as e:
logger.error("Failed while fetching replies {}".format(e))
break
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