我正在努力理解Python中的线程。我看过文档和示例,但坦率地说,许多示例过于复杂,我很难理解它们。
如何清楚地显示为多线程划分的任务?
我正在努力理解Python中的线程。我看过文档和示例,但坦率地说,许多示例过于复杂,我很难理解它们。
如何清楚地显示为多线程划分的任务?
当前回答
与其他提到的一样,由于GIL,CPython只能在I/O等待时使用线程。
如果您想从多个内核中获得CPU绑定任务的好处,请使用多处理:
from multiprocessing import Process
def f(name):
print 'hello', name
if __name__ == '__main__':
p = Process(target=f, args=('bob',))
p.start()
p.join()
其他回答
与其他提到的一样,由于GIL,CPython只能在I/O等待时使用线程。
如果您想从多个内核中获得CPU绑定任务的好处,请使用多处理:
from multiprocessing import Process
def f(name):
print 'hello', name
if __name__ == '__main__':
p = Process(target=f, args=('bob',))
p.start()
p.join()
Python 3具有启动并行任务的功能。这使我们的工作更容易。
它有线程池和进程池。
以下内容提供了一个见解:
ThreadPoolExecutor示例(源代码)
import concurrent.futures
import urllib.request
URLS = ['http://www.foxnews.com/',
'http://www.cnn.com/',
'http://europe.wsj.com/',
'http://www.bbc.co.uk/',
'http://some-made-up-domain.com/']
# Retrieve a single page and report the URL and contents
def load_url(url, timeout):
with urllib.request.urlopen(url, timeout=timeout) as conn:
return conn.read()
# We can use a with statement to ensure threads are cleaned up promptly
with concurrent.futures.ThreadPoolExecutor(max_workers=5) as executor:
# Start the load operations and mark each future with its URL
future_to_url = {executor.submit(load_url, url, 60): url for url in URLS}
for future in concurrent.futures.as_completed(future_to_url):
url = future_to_url[future]
try:
data = future.result()
except Exception as exc:
print('%r generated an exception: %s' % (url, exc))
else:
print('%r page is %d bytes' % (url, len(data)))
ProcessPoolExecutor(源)
import concurrent.futures
import math
PRIMES = [
112272535095293,
112582705942171,
112272535095293,
115280095190773,
115797848077099,
1099726899285419]
def is_prime(n):
if n % 2 == 0:
return False
sqrt_n = int(math.floor(math.sqrt(n)))
for i in range(3, sqrt_n + 1, 2):
if n % i == 0:
return False
return True
def main():
with concurrent.futures.ProcessPoolExecutor() as executor:
for number, prime in zip(PRIMES, executor.map(is_prime, PRIMES)):
print('%d is prime: %s' % (number, prime))
if __name__ == '__main__':
main()
作为第二个anwser的python3版本:
import queue as Queue
import threading
import urllib.request
# Called by each thread
def get_url(q, url):
q.put(urllib.request.urlopen(url).read())
theurls = ["http://google.com", "http://yahoo.com", "http://www.python.org","https://wiki.python.org/moin/"]
q = Queue.Queue()
def thread_func():
for u in theurls:
t = threading.Thread(target=get_url, args = (q,u))
t.daemon = True
t.start()
s = q.get()
def non_thread_func():
for u in theurls:
get_url(q,u)
s = q.get()
您可以测试它:
start = time.time()
thread_func()
end = time.time()
print(end - start)
start = time.time()
non_thread_func()
end = time.time()
print(end - start)
non_thread_func()花费的时间应该是thread_func()的4倍
对我来说,线程的最佳示例是监视异步事件。看看这个代码。
# thread_test.py
import threading
import time
class Monitor(threading.Thread):
def __init__(self, mon):
threading.Thread.__init__(self)
self.mon = mon
def run(self):
while True:
if self.mon[0] == 2:
print "Mon = 2"
self.mon[0] = 3;
您可以通过打开IPython会话并执行以下操作来使用此代码:
>>> from thread_test import Monitor
>>> a = [0]
>>> mon = Monitor(a)
>>> mon.start()
>>> a[0] = 2
Mon = 2
>>>a[0] = 2
Mon = 2
等几分钟
>>> a[0] = 2
Mon = 2
这里是多线程,有一个简单的例子会很有帮助。您可以运行它并轻松了解多线程在Python中的工作方式。我使用了一个锁来防止访问其他线程,直到前面的线程完成它们的工作。通过使用这行代码,
t锁定=线程。有界信号量(值=4)
您可以一次允许多个进程,并保留将在稍后或完成之前的进程后运行的其余线程。
import threading
import time
#tLock = threading.Lock()
tLock = threading.BoundedSemaphore(value=4)
def timer(name, delay, repeat):
print "\r\nTimer: ", name, " Started"
tLock.acquire()
print "\r\n", name, " has the acquired the lock"
while repeat > 0:
time.sleep(delay)
print "\r\n", name, ": ", str(time.ctime(time.time()))
repeat -= 1
print "\r\n", name, " is releaseing the lock"
tLock.release()
print "\r\nTimer: ", name, " Completed"
def Main():
t1 = threading.Thread(target=timer, args=("Timer1", 2, 5))
t2 = threading.Thread(target=timer, args=("Timer2", 3, 5))
t3 = threading.Thread(target=timer, args=("Timer3", 4, 5))
t4 = threading.Thread(target=timer, args=("Timer4", 5, 5))
t5 = threading.Thread(target=timer, args=("Timer5", 0.1, 5))
t1.start()
t2.start()
t3.start()
t4.start()
t5.start()
print "\r\nMain Complete"
if __name__ == "__main__":
Main()