对于Python中线程问题的简单讲解
我们将会看到一些在Python中使用线程的实例和如何避免线程之间的竞争。你应当将下边的例子运行多次,以便可以注意到线程是不可预测的和线程每次运行出的不同结果。声明:从这里开始忘掉你听到过的关于GIL的东西,因为GIL不会影响到我想要展示的东西。 示例1 我们将要请求五个不同的url: import time import urllib2 def get_responses(): urls = [ 'http://www.google.com','http://www.amazon.com','http://www.ebay.com','http://www.alibaba.com','http://www.reddit.com' ] start = time.time() for url in urls: print url resp = urllib2.urlopen(url) print resp.getcode() print "Elapsed time: %s" % (time.time()-start) get_responses() 输出是: http://www.google.com 200 http://www.amazon.com 200 http://www.ebay.com 200 http://www.alibaba.com 200 http://www.reddit.com 200 Elapsed time: 3.0814409256 解释:
多线程 import urllib2 import time from threading import Thread class GetUrlThread(Thread): def __init__(self,url): self.url = url super(GetUrlThread,self).__init__() def run(self): resp = urllib2.urlopen(self.url) print self.url,resp.getcode() def get_responses(): urls = [ 'http://www.google.com','http://www.reddit.com' ] start = time.time() threads = [] for url in urls: t = GetUrlThread(url) threads.append(t) t.start() for t in threads: t.join() print "Elapsed time: %s" % (time.time()-start) get_responses() 输出: http://www.reddit.com 200 http://www.google.com 200 http://www.amazon.com 200 http://www.alibaba.com 200 http://www.ebay.com 200 Elapsed time: 0.689890861511 解释:
关于线程:
实例2 我们将会用一个程序演示一下多线程间的资源竞争,并修复这个问题。 from threading import Thread #define a global variable some_var = 0 class IncrementThread(Thread): def run(self): #we want to read a global variable #and then increment it global some_var read_value = some_var print "some_var in %s is %d" % (self.name,read_value) some_var = read_value + 1 print "some_var in %s after increment is %d" % (self.name,some_var) def use_increment_thread(): threads = [] for i in range(50): t = IncrementThread() threads.append(t) t.start() for t in threads: t.join() print "After 50 modifications,some_var should have become 50" print "After 50 modifications,some_var is %d" % (some_var,) use_increment_thread() 多次运行这个程序,你会看到多种不同的结果。 解释:
为什么没有达到50?
解决资源竞争 from threading import Lock,Thread lock = Lock() some_var = 0 class IncrementThread(Thread): def run(self): #we want to read a global variable #and then increment it global some_var lock.acquire() read_value = some_var print "some_var in %s is %d" % (self.name,some_var) lock.release() def use_increment_thread(): threads = [] for i in range(50): t = IncrementThread() threads.append(t) t.start() for t in threads: t.join() print "After 50 modifications,) use_increment_thread() 再次运行这个程序,达到了我们预期的结果。 解释:
实例3 让我们用一个例子来证明一个线程不能影响其他线程内的变量(非全局变量)。 time.sleep()可以使一个线程挂起,强制线程切换发生。 from threading import Thread import time class CreateListThread(Thread): def run(self): self.entries = [] for i in range(10): time.sleep(1) self.entries.append(i) print self.entries def use_create_list_thread(): for i in range(3): t = CreateListThread() t.start() use_create_list_thread() 运行几次后发现并没有打印出争取的结果。当一个线程正在打印的时候,cpu切换到了另一个线程,所以产生了不正确的结果。我们需要确保print self.entries是个逻辑上的原子操作,以防打印时被其他线程打断。 我们使用了Lock(),来看下边的例子。 from threading import Thread,Lock import time lock = Lock() class CreateListThread(Thread): def run(self): self.entries = [] for i in range(10): time.sleep(1) self.entries.append(i) lock.acquire() print self.entries lock.release() def use_create_list_thread(): for i in range(3): t = CreateListThread() t.start() 这次我们看到了正确的结果。证明了一个线程不可以修改其他线程内部的变量(非全局变量)。 (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |