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我目前正在製作一個推特刮板,並且我想要獲得帶有多個hashtags的所有推文。問題是我每次嘗試通過第一個hashtag時都會收到429個錯誤。我試着睡覺的功能,但每次第二個哈希標籤出現,它不起作用。達到了Python Tweepy限額
import tweepy
import time
import json
from collections import defaultdict as dd
f = open("tokens.txt", 'r')
consumer_key = f.readline().strip()
consumer_secret = f.readline().strip()
app_key = f.readline().strip()
app_secret = f.readline().strip()
auth =tweepy.OAuthHandler(consumer_key, consumer_secret)
auth.set_access_token(app_key,app_secret)
api = tweepy.API(auth,wait_on_rate_limit=True,wait_on_rate_limit_notify=True)
usercount = dd(int)
userfollowers = dd(int)
mostretweets = dd(int)
mostfav = dd(int)
hashtag = ['#csforall','#equality']
for i in hashtag :
for status in tweepy.Cursor(api.search, q=i,since="2017-02-25",until="2017-02-28",lang="en").items():
parsed = status._json
usercount[parsed['user']['name'].encode("utf-8")]+=1
userfollowers[parsed['user']['name'].encode("utf-8")]= parsed['user']['followers_count']
mostretweets[parsed['text'].encode('utf-8')] = parsed['retweet_count']
mostfav[parsed['text'].encode('utf-8')] = parsed['favorite_count']
time.sleep(2)
sortcount = sorted(usercount.items(), key=lambda x: x[1], reverse =True)
top = sortcount[:1]
frequser=[]
for i in sortcount:
if i[1] == top:
frequser.append(i)
else:
break
print ("Top most frequent user: \n " + str(i[0])) +"\n"
followcount = sorted(userfollowers.items(), key=lambda x: x[1], reverse =True)
fol = followcount[:1]
freqfollow = []
for j in followcount:
if j[1] == fol:
freqfollow.append(i)
else:
break
print ("User with most followers: \n " + str(j[1]))
retweetcount = sorted(mostretweets.items(), key=lambda x: x[1], reverse = True)
ret = retweetcount[:1]
freqretweet =[]
for i in retweetcount:
if i[1] == ret:
freqretweet == ret
else:
break
print str(i[0])+"\n"
favcount = sorted(mostfav.items(), key=lambda x: x[1], reverse = True)
ret = favcount[:1]
freqfav =[]
for i in favcount:
if i[1] == ret:
freqfav == ret
else:
break
print str(i[0])+"\n"
不,它不起作用。 #equality有太多的推文,並收到429錯誤。 – MountainSon
它是否適用於它的兩個time.sleeps? –
那麼當我在#equality上運行睡眠時,有太多的平等推文,它需要太長的時間,所以它似乎永遠不會結束 – MountainSon