时间:2021-07-01 10:21:17 帮助过:6人阅读
是学习python 多线程的工作原理,及通过抓取400张图片这种IO密集型应用来查看多线程效率对比
import requests import urlparse import os import time import threading import Queue path = '/home/lidongwei/scrapy/owan_img_urls.txt' #path = '/home/lidongwei/scrapy/cc.txt' fetch_img_save_path = '/home/lidongwei/scrapy/owan_imgs/' # 读取保存再文件里面400个urls with open(path) as f : urls = f.readlines() urls = urls[:400] # 使用Queue来线程通信,因为队列是线程安全的(就是默认这个队列已经有锁) q = Queue.Queue() for url in urls: q.put(url) start = time.time() def fetch_img_func(q): while True: try: # 不阻塞的读取队列数据 url = q.get_nowait() i = q.qsize() except Exception, e: print e break; print 'Current Thread Name Runing %s ... 11' % threading.currentThread().name url = url.strip() img_path = urlparse.urlparse(url).path ext = os.path.splitext(img_path)[1] print 'handle %s pic... pic url %s ' % (i, url) res = requests.get(url, stream=True) if res.status_code == 200: save_img_path = '%s%s%s' % (fetch_img_save_path, i, ext) # 保存下载的图片 with open(save_img_path, 'wb') as fs: for chunk in res.iter_content(1024): fs.write(chunk) print 'save %s pic ' % i # 可以开多个线程测试不同效果 t1 = threading.Thread(target=fetch_img_func, args=(q, ), name="child_thread_1") #t2 = threading.Thread(target=fetch_img_func, args=(q, ), name="child_thread_2") #t3 = threading.Thread(target=fetch_img_func, args=(q, ), name="child_thread_3") #t4 = threading.Thread(target=fetch_img_func, args=(q, ), name="child_thread_4") t1.start() #t2.start() #t3.start() #t4.start() t1.join() #t2.join() #t3.join() #t4.join() end = time.time() print 'Done %s ' % (end-start)
实验结果
400图片
4线程 Done 12.443133831 3线程 Done 12.9201757908 2线程 Done 32.8628299236 1线程 Done 54.6115460396
总结
Python 自带GIL 大锁, 没有真正意义上的多线程并行执行。GIL 大锁会在线程阻塞的时候释放,此时等待的线程就可以激活工作,这样如此类推,大大提高IO阻塞型应用的效率。