直接从医生那里

加入((超时)) 等待线程终止。这将阻塞调用线程，直到调用join()方法的线程终止—正常终止或通过未处理的异常终止—或者直到可选超时发生。

这意味着生成t和d的主线程等待t完成，直到它完成。

根据程序使用的逻辑，您可能希望等到主线程结束后再继续执行。

文档中还提到:

一个线程可以被标记为“守护线程”。这个标志的意义在于，当只剩下守护线程时，整个Python程序将退出。

举个简单的例子:

def non_daemon():
    time.sleep(5)
    print 'Test non-daemon'

t = threading.Thread(name='non-daemon', target=non_daemon)

t.start()

最后是:

print 'Test one'
t.join()
print 'Test two'

这将输出:

Test one
Test non-daemon
Test two

在这里，主线程显式地等待t线程完成，直到它第二次调用print。

另一种情况是:

print 'Test one'
print 'Test two'
t.join()

我们将得到这样的输出:

Test one
Test two
Test non-daemon

在这里，我们在主线程中完成工作，然后等待t线程完成。在这种情况下，我们甚至可以删除显式连接t.join()，程序将隐式地等待t完成。

这里似乎误解了同步和异步处理之间的区别。

A thread is meant to execute a sub-procedure, most of the times on a "parallel" or "concurrent" fashion (depends on whether the device has multi-processors or not). But, what's the point on concurrency? For the most part, it's about improving performance of a process, by applying the idea of "divide and conquer". Have several threads (sub-processes) executing a "portion" of the whole process simultaneously, and then have a "final" step where all sub-processes results are combined (joined; hence the "join" method).

Of course, in order to achieve such gain on efficiency, the portions that are divided into threads, must be "mutually exclusive" (i.e., they don't share values to be updated... -- known in parallel computing as "critical section" -- ). If there is at least one value that is updated by two or more threads, then one has to wait for the other to "finish" its update, otherwise obtaining inconsistent results (i.e., two persons owning a bank account intend to withdraw certain amount of money in an ATM... if there won't be a proper mechanism that "locks" or "protects" the variable "balance" in both of the ATM devices, withdraws will completely screw-up the final value of the balance, causing obvious serious financial problem to the account owners).

那么，回到并行计算中线程的目的:让所有线程完成各自的部分，并使用“join”使它们“回到”主进程，这样每个单独的结果就会被“合并”成一个全局结果。

例子吗?有很多，但让我们列举几个解释清楚的:

Matrix multiplication: have each thread multiplying a vector of matrix A by the whole second matrix B, to obtain a vector of matrix C. At the end, have all resulting vestors put together to "display" (show) result: matrix C. In this example, although matrix B is used by all threads, no value of it is ever updated or modified (read-only). Summation, product of an array of massive numbers (an array of thousand of values, whether integer or float). Make threads to execute partial sums/products (say, if you have to sum 10K values, create 5 threads, each with 2K values); then with "join" make them return to the main process and sum individual results of all 5 threads. Theoretically, the process will do 2000 + 5 steps (2000 simultaneously in 5 threads, plus summation of final 5 sub-totals in the main process). In practice, though, how long do the 5 threads take to do its own 2000 numbers summation is completely variable as different factors get involved here (processor speed, electrical flow, or if it is a web service, network latency, and so on). However, the amount ot time invested would be in the "worst case", the amount of time the "slowest" thread takes, plus the final summation of 5 results step. Also, in practice, a thread that is meant to do 20% of the whole job, unlikely will take much longer than a single sequential process that would do 100% of the job (of course, it also depends on the size of the sample to be processed... the advantage won't be the same on a summation of 10K values, than summation of just 10 values with the same 5 threads... it's non-practicall, not worth it). Quick sort: We all know in general how quick sort works. However, there's a chance to improve it, if, say, we execute it in TWO threads: one that does the odd numbers and one that does the even ones. Then executes recursively and at some point it joins results of both threads and does a final quick sort in a fashion that will not require so many repetitions as numbers will be sufficiently ordered after the two threads did its initial job. That's a serios gain on performance with a quite big and unordered number of items. Chances are three threads can be used by doing some arrangement to the logic behind it, but its gain is really minimum and not worth to be programmed. However, two threads have a decent performance (time) gain.

因此，在python中使用“join”(或在其他“并发性”语言中使用“join”)具有重要的意义;但这在很大程度上取决于编程理解她/他想要“并行化”什么，以及她/他在将算法分割成需要并行化的正确步骤和需要在主进程中保留哪些步骤方面的熟练程度。这更像是一个“逻辑”思考的问题，而不是编程的“反模式”问题。

您可能会说:“使用join()有什么用?”实际上，这和“关闭文件有什么用，因为python和操作系统会在程序退出时为我关闭文件?”的答案是一样的。

这只是一个好的编程问题。您应该在代码中线程不应该再运行的地方join()您的线程，因为您必须确保线程的运行不会干扰您自己的代码，或者您希望在更大的系统中正确地运行。

您可能会说“我不希望我的代码延迟给出答案”，只是因为join()可能需要额外的时间。在某些情况下，这可能是完全有效的，但现在你需要考虑到你的代码“留下了粗糙的东西，让python和操作系统来清理”。如果您这样做是出于性能原因，我强烈建议您记录该行为。如果您正在构建其他人希望使用的库/包，则尤其如此。

除了性能原因之外，没有理由不使用join()，而且我认为您的代码不需要执行得那么好。

下面的例子演示了.join()操作:

import threading
import time

def threaded_worker():
    for r in range(10):
        print('Other: ', r)
        time.sleep(2)

thread_ = threading.Timer(1, threaded_worker)
thread_.daemon = True  # If the main thread is killed, this thread will be killed as well. 
thread_.start()

flag = True

for i in range(10):
    print('Main: ', i)
    time.sleep(2)
    if flag and i > 4:
        print(
            '''
            Threaded_worker() joined to the main thread. 
            Now we have a sequential behavior instead of concurrency.
            ''')
        thread_.join()
        flag = False

Out:

Main:  0
Other:  0
Main:  1
Other:  1
Main:  2
Other:  2
Main:  3
Other:  3
Main:  4
Other:  4
Main:  5
Other:  5

            Threaded_worker() joined to the main thread. 
            Now we have a sequential behavior instead of concurrency.
            
Other:  6
Other:  7
Other:  8
Other:  9
Main:  6
Main:  7
Main:  8
Main:  9

谢谢你的这篇文章——它也帮了我很多。

我今天学了一些关于.join()的知识。

这些线程并行运行:

d.start()
t.start()
d.join()
t.join()

这些顺序运行(不是我想要的):

d.start()
d.join()
t.start()
t.join()

特别是，我试图聪明和整洁:

class Kiki(threading.Thread):
    def __init__(self, time):
        super(Kiki, self).__init__()
        self.time = time
        self.start()
        self.join()

这个工作!但它是按顺序运行的。我可以把self.start()放在__ init __中，但不是self.join()。这必须在启动每个线程之后完成。

Join()是导致主线程等待线程完成的原因。否则，线程将自行运行。

因此，有一种方法可以将join()视为主线程上的“hold”——它在某种程度上解除线程的线程，并在主线程继续执行之前在主线程中顺序执行。它确保主线程向前移动之前线程已经完成。请注意，这意味着如果在调用join()之前线程已经完成，也没关系——当调用join()时，主线程会立即被释放。

事实上，我刚刚想到主线程会在d.t join()上等待，直到线程d结束，然后才移动到t.t join()。

事实上，为了更清楚地说明问题，请考虑以下代码:

import threading
import time

class Kiki(threading.Thread):
    def __init__(self, time):
        super(Kiki, self).__init__()
        self.time = time
        self.start()

    def run(self):
        print self.time, " seconds start!"
        for i in range(0,self.time):
            time.sleep(1)
            print "1 sec of ", self.time
        print self.time, " seconds finished!"


t1 = Kiki(3)
t2 = Kiki(2)
t3 = Kiki(1)
t1.join()
print "t1.join() finished"
t2.join()
print "t2.join() finished"
t3.join()
print "t3.join() finished"

它产生这样的输出(注意print语句是如何相互衔接的)。

$ python test_thread.py
32   seconds start! seconds start!1

 seconds start!
1 sec of  1
 1 sec of 1  seconds finished!
 21 sec of
3
1 sec of  3
1 sec of  2
2  seconds finished!
1 sec of  3
3  seconds finished!
t1.join() finished
t2.join() finished
t3.join() finished
$ 

t1.join()占用主线程。在t1.join()结束之前，所有三个线程都完成了，主线程继续执行打印，然后t2.join()，然后打印，然后t3.join()，然后打印。

修正的欢迎。我也是线程的新手。

(注意:如果你感兴趣的话，我正在为DrinkBot编写代码，我需要线程来并发地运行配料泵，而不是按顺序运行——这样就可以减少等待每种饮料的时间。)

join()方法

阻塞调用线程，直到调用join()方法的线程终止。

来源:http://docs.python.org/2/library/threading.html