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
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
其他回答
考虑使用风车,虽然风车可能不能做那么多线程。
您可以在5台机器上使用手卷Python脚本,每台机器使用端口40000-60000连接出站,打开100,000个端口连接。
另外,使用一个线程良好的QA应用程序(如OpenSTA)做一个示例测试可能会有所帮助,以了解每个服务器可以处理多少。
另外,试着在LWP::ConnCache类中使用简单的Perl。这样您可能会获得更好的性能(更多的连接)。
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。
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
(下一个项目的自我提示)
Python 3解决方案只使用请求。它是最简单且快速的,不需要多处理或复杂的异步库。
最重要的方面是重用连接,特别是对于HTTPS (TLS需要额外的往返才能打开)。注意,连接是特定于子域的。如果在多个域上抓取多个页面,则可以对url列表进行排序,以最大化连接重用(它有效地按域进行排序)。
当给定足够的线程时,它将与任何异步代码一样快。(请求在等待响应时释放python GIL)。
[带有日志记录和错误处理的生产等级代码]
import logging
import requests
import time
from concurrent.futures import ThreadPoolExecutor, as_completed
# source: https://stackoverflow.com/a/68583332/5994461
THREAD_POOL = 16
# This is how to create a reusable connection pool with python requests.
session = requests.Session()
session.mount(
'https://',
requests.adapters.HTTPAdapter(pool_maxsize=THREAD_POOL,
max_retries=3,
pool_block=True)
)
def get(url):
response = session.get(url)
logging.info("request was completed in %s seconds [%s]", response.elapsed.total_seconds(), response.url)
if response.status_code != 200:
logging.error("request failed, error code %s [%s]", response.status_code, response.url)
if 500 <= response.status_code < 600:
# server is overloaded? give it a break
time.sleep(5)
return response
def download(urls):
with ThreadPoolExecutor(max_workers=THREAD_POOL) as executor:
# wrap in a list() to wait for all requests to complete
for response in list(executor.map(get, urls)):
if response.status_code == 200:
print(response.content)
def main():
logging.basicConfig(
format='%(asctime)s.%(msecs)03d %(levelname)-8s %(message)s',
level=logging.INFO,
datefmt='%Y-%m-%d %H:%M:%S'
)
urls = [
"https://httpstat.us/200",
"https://httpstat.us/200",
"https://httpstat.us/200",
"https://httpstat.us/404",
"https://httpstat.us/503"
]
download(urls)
if __name__ == "__main__":
main()
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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