python - Linux允许python使用多少个网口？

Question

所以我一直在尝试在 python 中对一些互联网连接进行多线程处理。我一直在使用多处理模块，所以我可以绕过“全局解释器锁”。但是好像系统只给python开放了一个连接端口，或者至少一次只允许一个连接发生。这是我所说的示例。

*注意这是在linux服务器上运行

from multiprocessing import Process, Queue
import urllib
import random

# Generate 10,000 random urls to test and put them in the queue
queue = Queue()
for each in range(10000):
    rand_num = random.randint(1000,10000)
    url = ('http://www.' + str(rand_num) + '.com')
    queue.put(url)

# Main funtion for checking to see if generated url is active
def check(q):
    while True:
        try:
            url = q.get(False)
            try:
                request = urllib.urlopen(url)
                del request
                print url + ' is an active url!'
            except:
                print url + ' is not an active url!'
        except:
            if q.empty():
                break

# Then start all the threads (50)
for thread in range(50):
    task = Process(target=check, args=(queue,))
    task.start()

因此，如果您运行它，您会注意到它在该函数上启动了 50 个实例，但一次只运行一个。您可能认为“全局解释器锁”正在执行此操作，但事实并非如此。尝试将函数更改为数学函数而不是网络请求，您将看到所有五十个线程同时运行。

那么我必须使用套接字吗？或者我能做些什么来让 python 访问更多端口？或者有什么我没有看到的东西？让我知道你的想法!谢谢!

*编辑

所以我写了这个脚本来更好地测试请求库。好像我之前没有很好地测试过它。 (我主要用的是urllib和urllib2)

from multiprocessing import Process, Queue
from threading import Thread
from Queue import Queue as Q
import requests
import time

# A main timestamp
main_time = time.time()

# Generate 100 urls to test and put them in the queue
queue = Queue()
for each in range(100):
    url = ('http://www.' + str(each) + '.com')
    queue.put(url)

# Timer queue
time_queue = Queue()

# Main funtion for checking to see if generated url is active
def check(q, t_q): # args are queue and time_queue
    while True:
        try:
            url = q.get(False)
            # Make a timestamp
            t = time.time()
            try:
                request = requests.head(url, timeout=5)
                t = time.time() - t
                t_q.put(t)
                del request
            except:
                t = time.time() - t
                t_q.put(t)
        except:
            break

# Then start all the threads (20)
thread_list = []
for thread in range(20):
    task = Process(target=check, args=(queue, time_queue))
    task.start()
    thread_list.append(task)

# Join all the threads so the main process don't quit
for each in thread_list:
    each.join()
main_time_end = time.time()

# Put the timerQueue into a list to get the average
time_queue_list = []
while True:
    try:
        time_queue_list.append(time_queue.get(False))
    except:
        break

# Results of the time
average_response = sum(time_queue_list) / float(len(time_queue_list))
total_time = main_time_end - main_time
line =  "Multiprocessing: Average response time: %s sec. -- Total time: %s sec." % (average_response, total_time)
print line

# A main timestamp
main_time = time.time()

# Generate 100 urls to test and put them in the queue
queue = Q()
for each in range(100):
    url = ('http://www.' + str(each) + '.com')
    queue.put(url)

# Timer queue
time_queue = Queue()

# Main funtion for checking to see if generated url is active
def check(q, t_q): # args are queue and time_queue
    while True:
        try:
            url = q.get(False)
            # Make a timestamp
            t = time.time()
            try:
                request = requests.head(url, timeout=5)
                t = time.time() - t
                t_q.put(t)
                del request
            except:
                t = time.time() - t
                t_q.put(t)
        except:
            break

# Then start all the threads (20)
thread_list = []
for thread in range(20):
    task = Thread(target=check, args=(queue, time_queue))
    task.start()
    thread_list.append(task)

# Join all the threads so the main process don't quit
for each in thread_list:
    each.join()
main_time_end = time.time()

# Put the timerQueue into a list to get the average
time_queue_list = []
while True:
    try:
        time_queue_list.append(time_queue.get(False))
    except:
        break

# Results of the time
average_response = sum(time_queue_list) / float(len(time_queue_list))
total_time = main_time_end - main_time
line =  "Standard Threading: Average response time: %s sec. -- Total time: %s sec." % (average_response, total_time)
print line

# Do the same thing all over again but this time do each url at a time
# A main timestamp
main_time = time.time()

# Generate 100 urls and test them
timer_list = []
for each in range(100):
    url = ('http://www.' + str(each) + '.com')
    t = time.time()
    try:
        request = requests.head(url, timeout=5)
        timer_list.append(time.time() - t)
    except:
        timer_list.append(time.time() - t)
main_time_end = time.time()

# Results of the time
average_response = sum(timer_list) / float(len(timer_list))
total_time = main_time_end - main_time
line = "Not using threads: Average response time: %s sec. -- Total time: %s sec." % (average_response, total_time)
print line

如您所见，它非常适合多线程。实际上，我的大部分测试表明线程模块实际上比多处理模块更快。 (我不明白为什么!)这是我的一些结果。

Multiprocessing: Average response time: 2.40511314869 sec. -- Total time: 25.6876308918 sec.
Standard Threading: Average response time: 2.2179402256 sec. -- Total time: 24.2941861153 sec.
Not using threads: Average response time: 2.1740363431 sec. -- Total time: 217.404567957 sec.

这是在我的家庭网络上完成的，我服务器上的响应时间要快得多。我认为我的问题已经得到间接回答，因为我在一个更复杂的脚本上遇到了问题。所有的建议都帮助我很好地优化了它。感谢大家!

Answer 1

it starts 50 instances on the function but only runs one at a time

您误解了 htop 的结果。只有少数(如果有的话)python 副本可以在任何特定实例上运行。它们中的大多数将被阻塞等待网络 I/O。

事实上，这些进程是并行运行的。

Try changing the function to a mathematical function instead of a network request and you will see that all fifty threads run simultaneously.

将任务更改为数学函数仅说明了 CPU 绑定(bind)(例如数学)和 IO 绑定(bind)(例如 urlopen)进程之间的区别。前者始终可运行，后者很少可运行。

it only prints one at a time. If it was actually running multiple processes it would print many out at once.

它一次打印一个，因为您正在向终端写入行。因为这些行是无法区分的，所以您无法判断它们是全部由一个线程编写的，还是由一个单独的线程依次编写的。