这里采用生产者消费者模式来设计多线程,消费者负责解析网页并得到一个网页上所有图片的url,而消费者则负责下载图片到本地即进行IO操作,在这里设计了5个消费者以及五个生产者
import requests
from lxml import etree
import os
import re
from urllib import request
from queue import Queue
import threading
HEADRES = {
'User-Agent':
'Mozilla/5.0 (Linux; Android 6.0; Nexus 5 Build/MRA58N) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/75.0.3770.100 Mobile Safari/537.36'
}
class Producers(threading.Thread):
def __init__(self, page_queue, img_queue, *args, **kwargs):
super(Producers, self).__init__(*args, **kwargs)
self.pq = page_queue
self.iq = img_queue
def run(self):
while True:
if self.pq.empty():
break
url = self.pq.get()
self.parse_page(url)
def getHtml(self, url):
r = requests.get(url, headers=HEADRES)
r.encoding = r.apparent_encoding
return r.text
def parse_page(self, url):
text = self.getHtml(url)
html = etree.HTML(text)
imgs = html.xpath('//div[@class="page-content text-center"]//img[@class!="gif"]')
imgurls = []
alts = []
for img in imgs:
img_url = img.get('data-original')
alt = img.get('alt')
if img_url not in imgurls:
imgurls.append(img_url) # 爬取下来的url每个都有两份 处理一下 存在就不加入
if alt not in alts:
alts.append(alt)
for value in zip(imgurls, alts):
imgurl, alt = value
alt1 = re.sub(r'[\??\.,。!!*]', '', alt) # windows文件名不能有这些字符 re处理掉
suffix = os.path.splitext(imgurl)[1]
filename = alt1 + suffix
self.iq.put((imgurl, filename))
class Customer(threading.Thread):
def __init__(self, page_queue, img_queue, *args, **kwargs):
super(Customer, self).__init__(*args, **kwargs)
self.pq = page_queue
self.iq = img_queue
def run(self):
while True:
if self.pq.empty() and self.iq.empty():
break
imgurl, filename = self.iq.get()
request.urlretrieve(imgurl, 'images/' + filename)
print(filename+'下载完毕')
if __name__ == '__main__':
page_queue = Queue(100)
img_queue = Queue(1000)
for i in range(1, 50):
url = 'http://www.doutula.com/photo/list/?page='+str(i)
page_queue.put(url)
for x in range(5):
t = Producers(page_queue, img_queue)
t.start()
for x in range(5):
t = Customer(page_queue, img_queue)
t.start()
可以与单线程爬取速度作比较,可明显观察到性能的提升