返回剩余的tweepy请求次数

Question

更新：我正在尝试以下代码，以便读取一组ID并获取它们对应的名称。我想使用remain_search_limits来避免出现速率限制错误。

limits = api.rate_limit_status()
remain_search_limits = limits['resources']['search']['/search/tweets']['remaining']
stream = open('myfile','w')
ss     = open('userNames', 'w')
for ids in content:
try:
    limits = api.rate_limit_status()
    remain_search_limits = limits['resources']['search']['/search/tweets']['remaining']
    print 'you have', remain_search_limits, 'API calls remaining until next hour'
    if remain_search_limits < 2:
         dtcode = datetime.utcnow()
         unixtime = calendar.timegm(dtcode.utctimetuple())
         sleeptime = rate_limit_json.get('reset_time_in_seconds') - unixtime + 10
         print 'waiting ', sleeptime, 'seconds'
         sleep(sleeptime)
    else:
         user = api.get_user(ids) 
         stream.write(str(user.id)+"\n")
         ss.write(str(user.name)+"\n")
except (tweepy.TweepError) as e:
    print e 
    stream.close()
    ss.close()

每次打印remain_search_limits时，它都会返回180，直到遇到tweepError异常。

Answer 1

这个例子告诉你怎么查看还剩多少条推文。

print rate_limit_json["resources"]["search"]['/search/tweets']['remaining']
180

"resources" is the key you should be using to access the information inside.

如果你想更新这个值，可以把它放在一个循环里，在每次调用time.sleep()之后重新赋值。

把所有代码放在一个while循环里面，像这样：

limits = api.rate_limit_status()
remain_search_limits = limits['resources']['search']['/search/tweets']['remaining']
while remain_search_limits >2:
  limits = api.rate_limit_status()
  remain_search_limits = limits['resources']['search']['/search/tweets']['remaining']
else:
   dtcode = datetime.utcnow()
   unixtime = calendar.timegm(dtcode.utctimetuple())
   sleeptime = rate_limit_json.get('reset_time_in_seconds') - unixtime + 10
   print 'waiting ', sleeptime, 'seconds'
   sleep(sleeptime)

我没有测试过这段代码，但应该差不多符合你的需求。

你可能想在每次调用之间加个暂停，我对这个API不太熟悉，所以不太确定你具体在做什么。