如何让爬虫运行n次，停止并等待10分钟，然后再次运行n次并重复整个过程？

Question

我有一个网络爬虫：

import http.client
import time
import threading

class Worker(threading.Thread):                            # Booting crawler workers
    def __init__(self, base, ref, conn, urlKey, url):
        threading.Thread.__init__(self)
        self.base = base
        self.ref = ref
        self.conn = conn
        self.urlKey = urlKey
        self.url = url
        
    def run_aux(self):                                     # Getting web response and do stuff with it
        self.conn.request('GET', self.url)
        response = self.conn.getresponse()
        status = response.status
        if status == 200:
            data = str(response.read())
           
            if data.find('No he podido entenderte.') != -1:
                data = 'Err: Ruta no comprendida'
            else:
                pos = data.find('>Id: ') + 5
                if pos == 4:
                    data = 'Err: Tag >Id: no encontrado'
                else:
                    data = data[pos:pos+32]  
        else:
            data = 'Err: ' + str(status)
        response.close()
        self.base.saveRes(self.urlKey, data, self.ref)
        
    def run(self):          # Auxiliary function that allows the crawler to  continue running if any problem is found while connecting to the web site. It will continue trying to runtill connections are accepted
        try:
            self.run_aux()
        except Exception as e:
            self.run()

class MIdCrawler(object):                               # Boot and instantiate the crawler
    def __init__(self, site, nThreads, urlDict):
        self.res = {}
        self.nThreads = nThreads
        self.urlDict = urlDict.copy()
        self.urlKeys = list(self.urlDict.keys())
        self.conns = [http.client.HTTPSConnection(site) for _ in range(self.nThreads)]
        self.threads_launched = 0
        self.launch()

    def newThread(self, ref):          # How the threads are launched
        if self.threads_launched < len(self.urlKeys):
            urlKey = self.urlKeys[self.threads_launched]
            Worker(self, ref, self.conns[ref], urlKey, self.urlDict.get(urlKey)).start()
            self.threads_launched += 1

    def launch(self):                       # Create new threads
        for i in range(self.nThreads):
            self.newThread(i)
            
    def saveRes(self, urlKey, data, ref):           # Saving the results into a dict
        self.res[urlKey] = data
        #print('Dato guardado')
        print(urlKey)
        self.newThread(ref)
    
    def getRes(self):                                      # Return results
        while len(self.res) < len(self.urlKeys):
            time.sleep(0.5)
        return(self.res)

    def close(self):                          # Close working threads
        for i in range(self.nThreads):
            self.conns[i].close()

这个爬虫会从我的网络服务器中恢复某些数据，当我的服务器收到调用时，它会调用谷歌地图 api 来恢复一些路线数据。我必须拨打大约 3000 个电话，问题是我注意到当在短时间内从同一 IP 拨打一定数量的连续电话时，谷歌地图会停止接听电话。出现此红旗的呼叫次数不是固定的，但始终在 240 到 300 次连续呼叫之间。

这个爬虫允许指定我们想要并行启动的线程（调用）的数量，问题是当我到达 2xx-300 调用时，谷歌地图服务器停止响应我的请求。

我是这个网络抓取的新手，我希望有人告诉我如何修改这个爬虫来运行 200 个调用，等待十分钟，然后从它停止的地方继续。

这是爬虫接收到的一些示例数据：

['/testchat?text=quiero%20ir%20desde%20carrer%20cervantes%2C%201%2C%20sant%20andreu%20de%20la%20barca%2C%20hasta%20avinguda%20constituci%C3%B3%2C%2024%2C%20sant%20andreu%20de%20la%20barca',
 '/testchat?text=quiero%20ir%20desde%20General%20Mitre%20239%2C%20Barcelona%2C%20hasta%20Plaza%20Artos%2C%20Barcelona',
 '/testchat?text=quiero%20ir%20desde%20carrer%20Bon%20viatge%2C%20sant%20Joan%20Desp%C3%AD%2C%20hasta%20mare%20de%20deu%20la%20Merc%C3%A8%2C%20sant%20Joan%20Desp%C3%AD',
 '/testchat?text=quiero%20ir%20desde%20no%20recuerdo%2C%20hasta%20no%20recuerdo',
 '/testchat?text=quiero%20ir%20desde%20travessera%20de%20les%20corts%2C%20barcelona%2C%20hasta%20Avenida%20diagonal%2050%2C%20barcelona',
 '/testchat?text=quiero%20ir%20desde%20Confidencial%2C%20hasta%20Confdencial%2C%20Confidencial',
 '/testchat?text=quiero%20ir%20desde%20Paseo%20Zona%20Franca%20241%2C%20Barcelona%2C%20hasta%20Plaza%20Espa%C3%B1a%2C%20Barcelona',
 '/testchat?text=quiero%20ir%20desde%20Rbla.%20les%20bobiles%2014%2C%20martorell%2C%20hasta%20plaza%20de%20la%20vila%201%2C%20martorell',
 '/testchat?text=quiero%20ir%20desde%20Rambla%20de%20Catalu%C3%B1a%2086%2C%20Barcelona%2C%20hasta%20Padilla%20342%2C%20Barcelona',
 '/testchat?text=quiero%20ir%20desde%20Concepci%C3%B3n%2C%20Abrera%2C%20hasta%20Major%2C%20Abrera']

对爬虫的调用示例：

site = 'sema-dev-backend.mybluemix.net'

urlDict = df.url.reset_index(drop = True).to_dict()

nThreads = 5

ts1 = time.time()

mIdCrawler = MIdCrawler(site, nThreads, urlDict)
print(mIdCrawler.getRes())

t = time.time() - ts1

print('secs: ' + str(t))
print('mean_time: ' + str(t / len(urlDict))) 

我应该在这段代码中修改什么来对我的 Web 服务进行 200 次调用，停止并等待 10 分钟，然后再在停止的地方继续进行 200 次调用并重复该过程？

非常感谢您

Answer 1

如果您创建一个 main() 函数，我使用您提供的代码创建该函数：

def main()
    site = 'sema-dev-backend.mybluemix.net'
    urlDict = df.url.reset_index(drop = True).to_dict()
    nThreads = 5
    ts1 = time.time()
    mIdCrawler = MIdCrawler(site, nThreads, urlDict)
    print(mIdCrawler.getRes())
    t = time.time() - ts1
    print('secs: ' + str(t))
    print('mean_time: ' + str(t / len(urlDict))) 

然后你可以使用这个循环：

import time
try:
    while True:
        for i in range(0, 200): #execute 200 times
            main()
        time.sleep(60*10) #10 minute delay
except KeyboardInterrupt:
    print("stopping script")
    exit(0)