Python多线程爬虫适用于IO密集型，涉及到网络、磁盘IO的任务都是IO密集型任务，多线程可以明显提高效率，例如多线程爬虫，多线程文件处理等等。CPU密集型任务不适合使用多线程处理。

思路：将所有的url放到队列里面，在io非阻塞的情况下，利用线程从队列里面取数据，当处理完所有的url，直接退出

#!/usr/bin/env python
#-*- coding:utf-8 -*-
#env:python3.X

import threading, queue, time, requests
from  urllib.request import urlopen
from bs4 import BeautifulSoup
res = requests.get('http://land.fang.com/market/210100________1_0_1.html')
soup = BeautifulSoup(res.text,'html.parser')
urlQueue = queue.Queue()
for message in soup.select('.list28_text'):
    url = 'http://land.fang.com' + message.select('a')[0]['href']
    urlQueue.put(url)

def fetchUrl(urlQueue):
    while True:
        try:
            url = urlQueue.get_nowait()   #不阻塞的读取队列数据
            i = urlQueue.qsize()          #队列长度，取出一个长度就减少一个
        except Exception as e:
            break         #当取完的时候，退出循环
        #print ('Current Thread Name %s, Url: %s ' % (threading.currentThread().name, url))
        try:
            response = urlopen(url)
            responseCode = response.getcode()   #获取返回的状态码
        except Exception as e:
            continue
        if responseCode == 200:
            #抓取内容的数据处理可以放到这里
            detail = requests.get(url)
            soup1 = BeautifulSoup(detail.text, 'html.parser')
            messes = []
            for mess in soup1.select('.banbox tr td'):
                messes.append(mess.text)
            print(messes[1:3])
            #time.sleep(1)
if __name__ == '__main__':
    start = time.time()
    threads = []
    threadNum = 10
    for i in range(0, threadNum):
        t = threading.Thread(target=fetchUrl, args=(urlQueue,))
        threads.append(t)
        t.start()
    for t in threads:
        t.join()
    end = time.time()
    print ('the total time is: %s ' %  (end - start))