达到了Python Tweepy限额

Question

我目前正在制作一个推特刮板，并且我想要获得带有多个hashtags的所有推文。问题是我每次尝试通过第一个hashtag时都会收到429个错误。我试着睡觉的功能，但每次第二个哈希标签出现，它不起作用。

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" 

Answer 1

是否把这个：

for i in hashtag: 
    time.sleep(2) 
    for status in tweepy.Cursor(api.search, q=i,since="2017-02-25",until="2017-02-28",lang="en").items(): 

工作？