I am opening a file which has 100,000 URL's. I need to send an HTTP request to each URL and print the status code. I am using Python 2.6, and so far looked at the many confusing ways Python implements threading/concurrency. I have even looked at the python concurrence library, but cannot figure out how to write this program correctly. Has anyone come across a similar problem? I guess generally I need to know how to perform thousands of tasks in Python as fast as possible - I suppose that means 'concurrently'.
当前回答
使用线程池是一个很好的选择,这将使这相当容易。不幸的是,python并没有一个标准库来简化线程池。但这里有一个不错的图书馆,你应该开始: http://www.chrisarndt.de/projects/threadpool/
来自他们网站的代码示例:
pool = ThreadPool(poolsize)
requests = makeRequests(some_callable, list_of_args, callback)
[pool.putRequest(req) for req in requests]
pool.wait()
希望这能有所帮助。
其他回答
最简单的方法是使用Python的内置线程库。它们不是“真正的”/内核线程。它们有问题(比如序列化),但足够好了。你需要一个队列和线程池。这里有一个选项,但是编写自己的选项很简单。您无法并行处理所有100,000个调用,但可以同时发出100个(或左右)调用。
创建epoll对象, 打开许多客户端TCP套接字, 调整他们的发送缓冲区比请求头多一点, 发送一个请求头-它应该是即时的,只是放置到缓冲区, 在epoll对象中注册套接字 在epoll obect上做。poll, 从.poll中读取每个套接字的前3个字节, 将它们写入sys。Stdout后面跟着\n(不刷新), 关闭客户端套接字。
限制同时打开的套接字数量-在创建套接字时处理错误。只有当另一个套接字关闭时才创建新的套接字。 调整操作系统限制。 尝试分成几个(不是很多)进程:这可能有助于更有效地使用CPU。
一个解决方案:
from twisted.internet import reactor, threads
from urlparse import urlparse
import httplib
import itertools
concurrent = 200
finished=itertools.count(1)
reactor.suggestThreadPoolSize(concurrent)
def getStatus(ourl):
url = urlparse(ourl)
conn = httplib.HTTPConnection(url.netloc)
conn.request("HEAD", url.path)
res = conn.getresponse()
return res.status
def processResponse(response,url):
print response, url
processedOne()
def processError(error,url):
print "error", url#, error
processedOne()
def processedOne():
if finished.next()==added:
reactor.stop()
def addTask(url):
req = threads.deferToThread(getStatus, url)
req.addCallback(processResponse, url)
req.addErrback(processError, url)
added=0
for url in open('urllist.txt'):
added+=1
addTask(url.strip())
try:
reactor.run()
except KeyboardInterrupt:
reactor.stop()
Testtime:
[kalmi@ubi1:~] wc -l urllist.txt
10000 urllist.txt
[kalmi@ubi1:~] time python f.py > /dev/null
real 1m10.682s
user 0m16.020s
sys 0m10.330s
[kalmi@ubi1:~] head -n 6 urllist.txt
http://www.google.com
http://www.bix.hu
http://www.godaddy.com
http://www.google.com
http://www.bix.hu
http://www.godaddy.com
[kalmi@ubi1:~] python f.py | head -n 6
200 http://www.bix.hu
200 http://www.bix.hu
200 http://www.bix.hu
200 http://www.bix.hu
200 http://www.bix.hu
200 http://www.bix.hu
Pingtime:
bix.hu is ~10 ms away from me
godaddy.com: ~170 ms
google.com: ~30 ms
一个使用tornado的异步网络库解决方案
from tornado import ioloop, httpclient
i = 0
def handle_request(response):
print(response.code)
global i
i -= 1
if i == 0:
ioloop.IOLoop.instance().stop()
http_client = httpclient.AsyncHTTPClient()
for url in open('urls.txt'):
i += 1
http_client.fetch(url.strip(), handle_request, method='HEAD')
ioloop.IOLoop.instance().start()
这段代码使用非阻塞网络I/O,没有任何限制。它可以扩展到数万个打开的连接。它将在单个线程中运行,但比任何线程解决方案都要快。签出非阻塞I/O
我发现使用tornado包是最快和最简单的方法来实现这一点:
from tornado import ioloop, httpclient, gen
def main(urls):
"""
Asynchronously download the HTML contents of a list of URLs.
:param urls: A list of URLs to download.
:return: List of response objects, one for each URL.
"""
@gen.coroutine
def fetch_and_handle():
httpclient.AsyncHTTPClient.configure(None, defaults=dict(user_agent='MyUserAgent'))
http_client = httpclient.AsyncHTTPClient()
waiter = gen.WaitIterator(*[http_client.fetch(url, raise_error=False, method='HEAD')
for url in urls])
results = []
# Wait for the jobs to complete
while not waiter.done():
try:
response = yield waiter.next()
except httpclient.HTTPError as e:
print(f'Non-200 HTTP response returned: {e}')
continue
except Exception as e:
print(f'An unexpected error occurred querying: {e}')
continue
else:
print(f'URL \'{response.request.url}\' has status code <{response.code}>')
results.append(response)
return results
loop = ioloop.IOLoop.current()
web_pages = loop.run_sync(fetch_and_handle)
return web_pages
my_urls = ['url1.com', 'url2.com', 'url100000.com']
responses = main(my_urls)
print(responses[0])
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