Question

我想处理180个请求/ 15分钟的Search-API速率限制。我想出的第一个解决方案是检查标题中的剩余请求并等待900秒。请参阅以下代码段：

results = search_interface.cursor(search_interface.search, q=k, lang=lang, result_type=result_mode)

while True:
    try:
        tweet = next(results)
        if limit_reached(search_interface):
            sleep(900)

        self.writer(tweet)


def limit_reached(search_interface):
    remaining_rate = int(search_interface.get_lastfunction_header('X-Rate-Limit-Remaining'))
    return remaining_rate <= 2

但看起来，标题信息在达到剩余的两个请求后不会重置为180。

我提出的第二个解决方案是处理速率限制的twython异常并等待剩余的时间：

results = search_interface.cursor(search_interface.search, q=k, lang=lang, result_type=result_mode)
while True:
    try:
        tweet = next(results)

        self.writer(tweet)
    except TwythonError as inst:
        logger.error(inst.msg)
        wait_for_reset(search_interface)
        continue
    except StopIteration:
        break


def wait_for_reset(search_interface):
      reset_timestamp = int(search_interface.get_lastfunction_header('X-Rate-Limit-Reset'))
      now_timestamp = datetime.now().timestamp()
      seconds_offset = 10

      t = reset_timestamp - now_timestamp + seconds_offset
      logger.info('Waiting {0} seconds for Twitter rate limit reset.'.format(t))
      sleep(t)

但是通过这个解决方案，我收到了这条消息INFO：重置丢弃的连接：api.twitter.com“并且循环不会继续生成器的最后一个元素。有人遇到同样的问题吗？

的问候。

Answer 1

只是限制自己是我的建议（假设你不断达到极限......）

QUERY_PER_SEC = 15*60/180.0  #180 per 15 minutes
#~5 seconds per query
class TwitterBot:
    last_update=0
    def doQuery(self,*args,**kwargs):
        tdiff = time.time()-self.last_update
        if tdiff < QUERY_PER_SEC:
            time.sleep(QUERY_PER_SEC-tdiff) 
        self.last_update = time.time()
        return search_interface.cursor(*args,**kwargs)

twython搜索api速率限制：标题信息不会更新

1 个答案: