查询特定Twitter用户的tweepy(get_followers)中的分页

Question

我对使用游标类的tweepy和分页相当新.我一直在尝试使用光标类来获取特定Twitter用户的所有关注者,但我不断得到它所说的错误"tweepy.error.TweepError: This method does not perform pagination" 因此我真的很感激任何帮助,如果有人可以帮助我实现这个任务获得所有的追随者一个特殊的Twitter用户,使用tweepy进行分页.我到目前为止的代码如下:

import tweepy

consumer_key='xyz'
consumer_secret='xyz'

access_token='abc'
access_token_secret='def'

auth = tweepy.OAuthHandler(consumer_key, consumer_secret)

auth.set_access_token(access_token, access_token_secret)

api = tweepy.API(auth)


user = api.get_user('somehandle')
print user.name

followers = tweepy.Cursor(user.followers)
temp=[]
for user in followers.items():
    temp.append(user)
    print temp
#the following part works fine but that is without pagination so I will be able to retrieve at #most 100 followers
aDict =  user.followers()
for friend in aDict:
    friendDict = friend.__getstate__()
    print friendDict['screen_name']

Answer 1

有一种方便的方法叫做followers_ids.它为给定(或,或)返回最多5000个关注者(twitter api限制)ID .screen_nameiduser_idcursor

然后,您可以在python中手动分页这些结果并调用lookup_users每个块.只要一次lookup_users只能处理100个用户ID(twitter api限制),将块大小设置为100是非常合理的.

这是代码(分页部分取自这里):

import itertools
import tweepy


def paginate(iterable, page_size):
    while True:
        i1, i2 = itertools.tee(iterable)
        iterable, page = (itertools.islice(i1, page_size, None),
                list(itertools.islice(i2, page_size)))
        if len(page) == 0:
            break
        yield page


auth = tweepy.OAuthHandler(<consumer_key>, <consumer_secret>)
auth.set_access_token(<key>, <secret>)

api = tweepy.API(auth)

followers = api.followers_ids(screen_name='gvanrossum')

for page in paginate(followers, 100):
    results = api.lookup_users(user_ids=page)
    for result in results:
        print result.screen_name

希望有所帮助.