我试图获得公司的追随者人数,并随着时间的推移进行跟踪。我有超过20万家公司,因此,以当前的api限制运行,我目前拥有的代码实际上需要花费数年的时间。
c = tweepy.Cursor(api.followers_ids, id = a)
ids = []
for id in c.items():
time.sleep(0.01)
ids.append(id) '
在此代码中,每个关注者都可以找到一个api。我想知道是否有一个函数仅将跟随者计数作为数字?另外,Twitter API限制是多少?
最佳答案
每个API一次最多返回5000个关注者ID,以检索20万家公司的所有关注者,这是Matthew A. Russell的《挖掘社交网络》一书中非常有用的脚本,用于解决twitter api限制
为了发出可靠的Twitter请求并访问Twitter的API,Matthew定义了以下方法:
import sys
import time
from urllib2 import URLError
from httplib import BadStatusLine
import json
import twitter
def oauth_login():
CONSUMER_KEY = ''
CONSUMER_SECRET = ''
OAUTH_TOKEN = ''
OAUTH_TOKEN_SECRET = ''
auth = twitter.oauth.OAuth(OAUTH_TOKEN, OAUTH_TOKEN_SECRET,
CONSUMER_KEY, CONSUMER_SECRET)
twitter_api = twitter.Twitter(auth=auth)
return twitter_api
def make_twitter_request(twitter_api_func, max_errors=10, *args, **kw):
# A nested helper function that handles common HTTPErrors. Return an updated
# value for wait_period if the problem is a 500 level error. Block until the
# rate limit is reset if it's a rate limiting issue (429 error). Returns None
# for 401 and 404 errors, which requires special handling by the caller.
def handle_twitter_http_error(e, wait_period=2, sleep_when_rate_limited=True):
if wait_period > 3600: # Seconds
print >> sys.stderr, 'Too many retries. Quitting.'
raise e
# See https://dev.twitter.com/docs/error-codes-responses for common codes
if e.e.code == 401:
print >> sys.stderr, 'Encountered 401 Error (Not Authorized)'
return None
elif e.e.code == 404:
print >> sys.stderr, 'Encountered 404 Error (Not Found)'
return None
elif e.e.code == 429:
print >> sys.stderr, 'Encountered 429 Error (Rate Limit Exceeded)'
if sleep_when_rate_limited:
print >> sys.stderr, "Retrying in 15 minutes...ZzZ..."
sys.stderr.flush()
time.sleep(60*15 + 5)
print >> sys.stderr, '...ZzZ...Awake now and trying again.'
return 2
else:
raise e # Caller must handle the rate limiting issue
elif e.e.code in (500, 502, 503, 504):
print >> sys.stderr, 'Encountered %iError. Retrying in %iseconds' %\
(e.e.code, wait_period)
time.sleep(wait_period)
wait_period *= 1.5
return wait_period
else:
raise e
# End of nested helper function
wait_period = 2
error_count = 0
while True:
try:
return twitter_api_func(*args, **kw)
except twitter.api.TwitterHTTPError, e:
error_count = 0
wait_period = handle_twitter_http_error(e, wait_period)
if wait_period is None:
return
except URLError, e:
error_count += 1
print >> sys.stderr, "URLError encountered. Continuing."
if error_count > max_errors:
print >> sys.stderr, "Too many consecutive errors...bailing out."
raise
except BadStatusLine, e:
error_count += 1
print >> sys.stderr, "BadStatusLine encountered. Continuing."
if error_count > max_errors:
print >> sys.stderr, "Too many consecutive errors...bailing out."
raise
这是检索朋友和关注者的方法:
from functools import partial
from sys import maxint
def get_friends_followers_ids(twitter_api, screen_name=None, user_id=None,
friends_limit=maxint, followers_limit=maxint):
# Must have either screen_name or user_id (logical xor)
assert (screen_name != None) != (user_id != None),\
"Must have screen_name or user_id, but not both"
# See https://dev.twitter.com/docs/api/1.1/get/friends/ids and
# https://dev.twitter.com/docs/api/1.1/get/followers/ids for details
# on API parameters
get_friends_ids = partial(make_twitter_request, twitter_api.friends.ids,
count=5000)
get_followers_ids = partial(make_twitter_request,twitter_api.followers.ids,
count=5000)
friends_ids, followers_ids = [], []
for twitter_api_func, limit, ids, label in [
[get_friends_ids, friends_limit, friends_ids, "friends"],
[get_followers_ids, followers_limit, followers_ids, "followers"]
]:
if limit == 0: continue
cursor = -1
while cursor != 0:
# Use make_twitter_request via the partially bound callable...
if screen_name:
response = twitter_api_func(screen_name=screen_name, cursor=cursor)
else: # user_id
response = twitter_api_func(user_id=user_id, cursor=cursor)
if response is not None:
ids += response['ids']
cursor = response['next_cursor']
print >> sys.stderr, 'Fetched {0} total {1} ids for{2}'.format(len(ids),
label, (user_id or screen_name))
# XXX: You may want to store data during each iteration to provide
# an additional layer of protection from exceptional circumstances
if len(ids) >= limit or response is None:
break
# Do something useful with the IDs, like store them to disk...
return friends_ids[:friends_limit], followers_ids[:followers_limit]
# Sample usage
twitter_api = oauth_login()
friends_ids, followers_ids =get_friends_followers_ids(twitter_api,
screen_name="SocialWebMining",
friends_limit=10,
followers_limit=10)
print friends_ids
print followers_ids
关于python - 如何使用tweepy获取关注者人数,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/29858708/