在 csv 文件中写入从推文中提取的 url

Question

我正在尝试使用此代码从推文中提取网址。它完美地工作，并将完整的网址作为输出提供给我。我想将所有这些 url 写在一个 csv 文件中，我不在乎它是否与推文相同（它会更好）或另一个。我尝试了不同的东西，比如“.to_csv”或 writerow 函数，但它们没有用，可能是因为我把它们放在了错误的地方。任何帮助表示赞赏！

def get_tweets(handle):
      try:
        auth = tweepy.OAuthHandler(consumer_key, consumer_secret)
        auth.set_access_token(access_token, access_token_secret)
        api = tweepy.API(auth)

    
        number_of_tweets = 200

        tweets = api.user_timeline(screen_name = handle,count = number_of_tweets)

        print(handle, "Number of tweets extracted: {}\n".format(len(tweets)))
        df = pd.DataFrame(data=[tweet.user.screen_name for tweet in tweets], columns=['ScreenName'])

        for tweet in tweets:
            urls = re.findall("http[s]?://(?:[a-zA-Z]|[0-9]|[$-_@.&+]|[!*\(\),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+", tweet.text)
            for url in urls:
                try:
                    opener = urllib.request.build_opener()
                    request = urllib.request.Request(url)
                    response = opener.open(request)
                    actual_url = response.geturl()
                    print (actual_url)
                except:
                    print(url)
      except:
        pass
      return df



handles = ["name of the user"]
           
         
for handle in handles:
    df_new = get_tweets(handle)

Answer 1

你可以试试这个：

handles = ["name of the user"]

for handle in handles:
    df_new = get_tweets(handle)
    df_new.to_csv(path_or_buf=f"{handle}_tweets.csv", index=False)