我想创建一个运行多个轻量级线程的程序，但是将其自身限制为一个恒定的，预定义数量的并发运行任务，就像这样（但没有竞争条件的风险）：
import threading
def f(arg):
global running
running += 1
print("Spawned a thread. running=%s, arg=%s" % (running, arg))
for i in range(100000):
pass
running -= 1
print("Done")
running = 0
while True:
if running < 8:
arg = get_task()
threading.Thread(target=f, args=[arg]).start()
实施此方法最安全/最快的方法是什么？
答案 0 :(得分：32)
听起来你想要用八个工人来实现生产者/消费者模式。为此目的，Python有一个Queue类，它是线程安全的。
每个工作人员都应该在队列上调用get()来检索任务。如果没有任何可用任务，此调用将阻止，从而导致工作程序空闲，直到有一个可用。然后工作人员应该执行任务，最后在队列上调用task_done()。
您可以通过调用队列中的put()将任务放入队列中。
在主线程中，您可以在队列上调用join()，等待所有挂起的任务完成。
这种方法的好处是您不会创建和销毁线程，这是昂贵的。工作线程将连续运行，但在没有任务进入队列时将使用零CPU时间处于睡眠状态。
（链接的文档页面有一个这种模式的例子。）
答案 1 :(得分：17)
semaphore是一种变量或抽象数据类型，用于控制并发系统（如多道程序设计操作系统）中多个进程对公共资源的访问;这可以帮到你。
threadLimiter = threading.BoundedSemaphore(maximumNumberOfThreads)
class MyThread(threading.Thread):
def run(self):
threadLimiter.acquire()
try:
self.Executemycode()
finally:
threadLimiter.release()
def Executemycode(self):
print(" Hello World!")
# <your code here>
这样，您可以轻松地限制在程序执行期间将同时执行的线程数。变量，＆＃39; maximumNumberOfThreads＆＃39;可用于定义线程最大值的上限。
答案 2 :(得分：5)
使用multiprocessing.dummy.Pool或concurrent.futures.ThreadPoolExecutor（或者，如果使用Python 2.x，后端futures）将此实现为线程池或执行器会容易得多。例如：
import concurrent
def f(arg):
print("Started a task. running=%s, arg=%s" % (running, arg))
for i in range(100000):
pass
print("Done")
with concurrent.futures.ThreadPoolExecutor(8) as executor:
while True:
arg = get_task()
executor.submit(f, arg)
当然，如果您可以将拉模型get_task更改为推模型get_tasks，例如，一次生成一个任务，则更简单：
with concurrent.futures.ThreadPoolExecutor(8) as executor:
for arg in get_tasks():
executor.submit(f, arg)
当你的任务用完时（例如get_task引发异常，或get_tasks运行干涸），这将自动告诉执行者在排空队列后停止，等待它停止，并清理一切。
答案 3 :(得分：3)
我见过最常见的写作：
threads = [threading.Thread(target=f) for _ in range(8)]
for thread in threads:
thread.start()
...
for thread in threads:
thread.join()
如果要维护一个固定大小的正在运行的线程池来处理短期任务而不是要求新工作，请考虑围绕队列构建的解决方案，例如“How to wait until only the first thread is finished in Python”。
答案 4 :(得分：1)
我遇到了同样的问题，花了几天（准确地说是2天）来使用队列找到正确的解决方案。我浪费了ThreadPoolExecutor的时间，因为没有办法限制事物启动的线程数量！我给它提供了一个要复制的5000个文件的列表，一旦它一次运行了大约1500个并发文件副本，代码就没有响应了。 ThreadPoolExecutor上的max_workers参数仅控制正在旋转线程的工人数，而不控制旋转的线程数。
好吧，无论如何，这是一个使用Queue的非常简单的示例：
pip install numpy==1.15.4
答案 5 :(得分：1)
concurrent.futures.ThreadPoolExecutor.map
concurrent.futures.ThreadPoolExecutor在https://stackoverflow.com/a/19370282/895245中被提及，这是map方法的示例，它通常是最方便的方法。
.map()是map()的并行版本：它立即读取所有输入，然后并行运行任务，并以与输入相同的顺序返回。
用法：
./concurrent_map_exception.py [nproc [min [max]]
concurrent_map_exception.py
import concurrent.futures
import sys
import time
def my_func(i):
time.sleep((abs(i) % 4) / 10.0)
return 10.0 / i
def my_get_work(min_, max_):
for i in range(min_, max_):
print('my_get_work: {}'.format(i))
yield i
# CLI.
argv_len = len(sys.argv)
if argv_len > 1:
nthreads = int(sys.argv[1])
if nthreads == 0:
nthreads = None
else:
nthreads = None
if argv_len > 2:
min_ = int(sys.argv[2])
else:
min_ = 1
if argv_len > 3:
max_ = int(sys.argv[3])
else:
max_ = 100
# Action.
with concurrent.futures.ProcessPoolExecutor(max_workers=nthreads) as executor:
for input, output in zip(
my_get_work(min_, max_),
executor.map(my_func, my_get_work(min_, max_))
):
print('result: {} {}'.format(input, output))
例如：
./concurrent_map_exception.py 1 1 5
给予：
my_get_work: 1
my_get_work: 2
my_get_work: 3
my_get_work: 4
my_get_work: 1
result: 1 10.0
my_get_work: 2
result: 2 5.0
my_get_work: 3
result: 3 3.3333333333333335
my_get_work: 4
result: 4 2.5
和：
./concurrent_map_exception.py 2 1 5
给出相同的输出，但运行速度更快，因为我们现在有2个进程，并且：
./concurrent_map_exception.py 1 -5 5
给予：
my_get_work: -5
my_get_work: -4
my_get_work: -3
my_get_work: -2
my_get_work: -1
my_get_work: 0
my_get_work: 1
my_get_work: 2
my_get_work: 3
my_get_work: 4
my_get_work: -5
result: -5 -2.0
my_get_work: -4
result: -4 -2.5
my_get_work: -3
result: -3 -3.3333333333333335
my_get_work: -2
result: -2 -5.0
my_get_work: -1
result: -1 -10.0
my_get_work: 0
concurrent.futures.process._RemoteTraceback:
"""
Traceback (most recent call last):
File "/usr/lib/python3.6/concurrent/futures/process.py", line 175, in _process_worker
r = call_item.fn(*call_item.args, **call_item.kwargs)
File "/usr/lib/python3.6/concurrent/futures/process.py", line 153, in _process_chunk
return [fn(*args) for args in chunk]
File "/usr/lib/python3.6/concurrent/futures/process.py", line 153, in <listcomp>
return [fn(*args) for args in chunk]
File "./concurrent_map_exception.py", line 24, in my_func
return 10.0 / i
ZeroDivisionError: float division by zero
"""
The above exception was the direct cause of the following exception:
Traceback (most recent call last):
File "./concurrent_map_exception.py", line 52, in <module>
executor.map(my_func, my_get_work(min_, max_))
File "/usr/lib/python3.6/concurrent/futures/process.py", line 366, in _chain_from_iterable_of_lists
for element in iterable:
File "/usr/lib/python3.6/concurrent/futures/_base.py", line 586, in result_iterator
yield fs.pop().result()
File "/usr/lib/python3.6/concurrent/futures/_base.py", line 432, in result
return self.__get_result()
File "/usr/lib/python3.6/concurrent/futures/_base.py", line 384, in __get_result
raise self._exception
ZeroDivisionError: float division by zero
所以请注意它如何在发生异常时立即停止。
Queue带有错误处理的示例
Queue，但这是一个完整的示例。
设计目标：
concurrent.futures.ThreadPoolExecutor是我见过的stdlib中当前可用的最好的接口。但是我找不到以下所有方法：
因为：
.map()：一次读取所有输入，func只能接受参数.submit()：.shutdown()一直执行到所有期货完成为止，并且对当前最大工作项目没有阻塞.submit()。那么如何避免第一次失败后所有期货上的丑陋的.cancel()循环呢？事不宜迟，这是我的实现。测试用例位于脚本的__name__ == '__main__'下：
thread_pool.py
#!/usr/bin/env python3
'''
This file is MIT Licensed because I'm posting it on Stack Overflow:
https://stackoverflow.com/questions/19369724/the-right-way-to-limit-maximum-number-of-threads-running-at-once/55263676#55263676
'''
from typing import Any, Callable, Dict, Iterable, Union
import os
import queue
import sys
import threading
import time
import traceback
class ThreadPoolExitException(Exception):
'''
An object of this class may be raised by output_handler_function to
request early termination.
It is also raised by submit() if submit_raise_exit=True.
'''
pass
class ThreadPool:
'''
Start a pool of a limited number of threads to do some work.
This is similar to the stdlib concurrent, but I could not find
how to reach all my design goals with that implementation:
* the input function does not need to be modified
* limit the number of threads
* queue sizes closely follow number of threads
* if an exception happens, optionally stop soon afterwards
This class form allows to use your own while loops with submit().
Exit soon after the first failure happens:
....
python3 thread_pool.py 2 -10 20 handle_output_print
....
Sample output:
....
{'i': -9} -1.1111111111111112 None
{'i': -8} -1.25 None
{'i': -10} -1.0 None
{'i': -6} -1.6666666666666667 None
{'i': -7} -1.4285714285714286 None
{'i': -4} -2.5 None
{'i': -5} -2.0 None
{'i': -2} -5.0 None
{'i': -3} -3.3333333333333335 None
{'i': 0} None ZeroDivisionError('float division by zero')
{'i': -1} -10.0 None
{'i': 1} 10.0 None
{'i': 2} 5.0 None
work_function or handle_output raised:
Traceback (most recent call last):
File "thread_pool.py", line 181, in _func_runner
work_function_return = self.work_function(**work_function_input)
File "thread_pool.py", line 281, in work_function_maybe_raise
return 10.0 / i
ZeroDivisionError: float division by zero
work_function_input: {'i': 0}
work_function_return: None
....
Don't exit after first failure, run until end:
....
python3 thread_pool.py 2 -10 20 handle_output_print_no_exit
....
Store results in a queue for later inspection instead of printing immediately,
then print everything at the end:
....
python3 thread_pool.py 2 -10 20 handle_output_queue
....
Exit soon after the handle_output raise.
....
python3 thread_pool.py 2 -10 20 handle_output_raise
....
Relying on this interface to abort execution is discouraged, this should
usually only happen due to a programming error in the handler.
Test that the argument called "thread_id" is passed to work_function and printed:
....
python3 thread_pool.py 2 -10 20 handle_output_print thread_id
....
Test with, ThreadPoolExitException and submit_raise_exit=True, same behaviour handle_output_print
except for the different exit cause report:
....
python3 thread_pool.py 2 -10 20 handle_output_raise_exit_exception
....
'''
def __init__(
self,
work_function: Callable,
handle_output: Union[Callable[[Any,Any,Exception],Any],None] = None,
nthreads: Union[int,None] = None,
thread_id_arg: Union[str,None] = None,
submit_raise_exit: bool = False
):
'''
Start in a thread pool immediately.
join() must be called afterwards at some point.
:param work_function: main work function to be evaluated.
:param handle_output: called on work_function return values as they
are returned.
The function signature is:
....
handle_output(
work_function_input: Union[Dict,None],
work_function_return,
work_function_exception: Exception
) -> Union[Exception,None]
....
where work_function_exception the exception that work_function raised,
or None otherwise
The first non-None return value of a call to this function is returned by
submit(), get_handle_output_result() and join().
The intended semantic for this, is to return:
* on success:
** None to continue execution
** ThreadPoolExitException() to request stop execution
* if work_function_input or work_function_exception raise:
** the exception raised
The ThreadPool user can then optionally terminate execution early on error
or request with either:
* an explicit submit() return value check + break if a submit loop is used
* `with` + submit_raise_exit=True
Default: a handler that just returns `exception`, which can normally be used
by the submit loop to detect an error and exit immediately.
:param nthreads: number of threads to use. Default: nproc.
:param thread_id_arg: if not None, set the argument of work_function with this name
to a 0-indexed thread ID. This allows function calls to coordinate
usage of external resources such as files or ports.
:param submit_raise_exit: if True, submit() raises ThreadPoolExitException() if
get_handle_output_result() is not None.
'''
self.work_function = work_function
if handle_output is None:
handle_output = lambda input, output, exception: exception
self.handle_output = handle_output
if nthreads is None:
nthreads = len(os.sched_getaffinity(0))
self.thread_id_arg = thread_id_arg
self.submit_raise_exit = submit_raise_exit
self.nthreads = nthreads
self.handle_output_result = None
self.handle_output_result_lock = threading.Lock()
self.in_queue = queue.Queue(maxsize=nthreads)
self.threads = []
for i in range(self.nthreads):
thread = threading.Thread(
target=self._func_runner,
args=(i,)
)
self.threads.append(thread)
thread.start()
def __enter__(self):
'''
__exit__ automatically calls join() for you.
This is cool because it automatically ends the loop if an exception occurs.
But don't forget that errors may happen after the last submit was called, so you
likely want to check for that with get_handle_output_result() after the with.
'''
return self
def __exit__(self, exception_type, exception_value, exception_traceback):
self.join()
return exception_type is ThreadPoolExitException
def _func_runner(self, thread_id):
while True:
work_function_input = self.in_queue.get(block=True)
if work_function_input is None:
break
if self.thread_id_arg is not None:
work_function_input[self.thread_id_arg] = thread_id
try:
work_function_exception = None
work_function_return = self.work_function(**work_function_input)
except Exception as e:
work_function_exception = e
work_function_return = None
handle_output_exception = None
try:
handle_output_return = self.handle_output(
work_function_input,
work_function_return,
work_function_exception
)
except Exception as e:
handle_output_exception = e
handle_output_result = None
if handle_output_exception is not None:
handle_output_result = handle_output_exception
elif handle_output_return is not None:
handle_output_result = handle_output_return
if handle_output_result is not None and self.handle_output_result is None:
with self.handle_output_result_lock:
self.handle_output_result = (
work_function_input,
work_function_return,
handle_output_result
)
self.in_queue.task_done()
@staticmethod
def exception_traceback_string(exception):
'''
Helper to get the traceback from an exception object.
This is usually what you want to print if an error happens in a thread:
https://stackoverflow.com/questions/3702675/how-to-print-the-full-traceback-without-halting-the-program/56199295#56199295
'''
return ''.join(traceback.format_exception(
None, exception, exception.__traceback__)
)
def get_handle_output_result(self):
'''
:return: if a handle_output call has raised previously, return a tuple:
....
(work_function_input, work_function_return, exception_raised)
....
corresponding to the first such raise.
Otherwise, if a handle_output returned non-None, a tuple:
(work_function_input, work_function_return, handle_output_return)
Otherwise, None.
'''
return self.handle_output_result
def join(self):
'''
Request all threads to stop after they finish currently submitted work.
:return: same as get_handle_output_result()
'''
for thread in range(self.nthreads):
self.in_queue.put(None)
for thread in self.threads:
thread.join()
return self.get_handle_output_result()
def submit(
self,
work_function_input: Union[Dict,None] =None
):
'''
Submit work. Block if there is already enough work scheduled (~nthreads).
:return: the same as get_handle_output_result
'''
handle_output_result = self.get_handle_output_result()
if handle_output_result is not None and self.submit_raise_exit:
raise ThreadPoolExitException()
if work_function_input is None:
work_function_input = {}
self.in_queue.put(work_function_input)
return handle_output_result
if __name__ == '__main__':
def get_work(min_, max_):
'''
Generate simple range work for work_function.
'''
for i in range(min_, max_):
yield {'i': i}
def work_function_maybe_raise(i):
'''
The main function that will be evaluated.
It sleeps to simulate an IO operation.
'''
time.sleep((abs(i) % 4) / 10.0)
return 10.0 / i
def work_function_get_thread(i, thread_id):
time.sleep((abs(i) % 4) / 10.0)
return thread_id
def handle_output_print(input, output, exception):
'''
Print outputs and exit immediately on failure.
'''
print('{!r} {!r} {!r}'.format(input, output, exception))
return exception
def handle_output_print_no_exit(input, output, exception):
'''
Print outputs, don't exit on failure.
'''
print('{!r} {!r} {!r}'.format(input, output, exception))
out_queue = queue.Queue()
def handle_output_queue(input, output, exception):
'''
Store outputs in a queue for later usage.
'''
global out_queue
out_queue.put((input, output, exception))
return exception
def handle_output_raise(input, output, exception):
'''
Raise if input == 0, to test that execution
stops nicely if this raises.
'''
print('{!r} {!r} {!r}'.format(input, output, exception))
if input['i'] == 0:
raise Exception
def handle_output_raise_exit_exception(input, output, exception):
'''
Return a ThreadPoolExitException() if input == -5.
Return the work_function exception if it raised.
'''
print('{!r} {!r} {!r}'.format(input, output, exception))
if exception:
return exception
if output == 10.0 / -5:
return ThreadPoolExitException()
# CLI arguments.
argv_len = len(sys.argv)
if argv_len > 1:
nthreads = int(sys.argv[1])
if nthreads == 0:
nthreads = None
else:
nthreads = None
if argv_len > 2:
min_ = int(sys.argv[2])
else:
min_ = 1
if argv_len > 3:
max_ = int(sys.argv[3])
else:
max_ = 100
if argv_len > 4:
handle_output_funtion_string = sys.argv[4]
else:
handle_output_funtion_string = 'handle_output_print'
handle_output = eval(handle_output_funtion_string)
if argv_len > 5:
work_function = work_function_get_thread
thread_id_arg = sys.argv[5]
else:
work_function = work_function_maybe_raise
thread_id_arg = None
# Action.
if handle_output is handle_output_raise_exit_exception:
# `with` version with implicit join and submit raise
# immediately when desired with ThreadPoolExitException.
#
# This is the more safe and convenient and DRY usage if
# you can use `with`, so prefer it generally.
with ThreadPool(
work_function,
handle_output,
nthreads,
thread_id_arg,
submit_raise_exit=True
) as my_thread_pool:
for work in get_work(min_, max_):
my_thread_pool.submit(work)
handle_output_result = my_thread_pool.get_handle_output_result()
else:
# Explicit error checking in submit loop to exit immediately
# on error.
my_thread_pool = ThreadPool(
work_function,
handle_output,
nthreads,
thread_id_arg,
)
for work_function_input in get_work(min_, max_):
handle_output_result = my_thread_pool.submit(work_function_input)
if handle_output_result is not None:
break
handle_output_result = my_thread_pool.join()
if handle_output_result is not None:
work_function_input, work_function_return, exception = handle_output_result
if type(exception) is ThreadPoolExitException:
print('Early exit requested by handle_output with ThreadPoolExitException:')
else:
print('work_function or handle_output raised:')
print(ThreadPool.exception_traceback_string(exception), end='')
print('work_function_input: {!r}'.format(work_function_input))
print('work_function_return: {!r}'.format(work_function_return))
if handle_output == handle_output_queue:
while not out_queue.empty():
print(out_queue.get())
在Python 3.7.3中进行了测试。
答案 6 :(得分：0)
这可以通过 Semaphore Object 来完成。信号量管理一个内部计数器，该计数器在每次 acquire() 调用时递减，在每次 release() 调用时递增。计数器永远不会低于零；当 acquire() 发现它为零时，它会阻塞，等待其他线程调用 release()。
一个简短的例子显示最多 5 个线程并行，其中一半线程立即执行，其他线程被阻塞并等待：
import threading
import time
maxthreads = 5
pool_sema = threading.Semaphore(value=maxthreads)
threads = list()
def task(i):
pool_sema.acquire()
try:
print("executed {}. thread".format(i))
time.sleep(2)
except Exception as e:
print("Error: problem with {0}. thread.\nMessage:{1}".format(i, e))
finally:
pool_sema.release()
def create_threads(number_of_threads):
try:
for i in range(number_of_threads):
thread = threading.Thread(target=task,args=(str(i)))
threads.append(thread)
thread.start()
except Exception as e:
print("Error: unable to start thread {}".format(e))
if __name__ == '__main__':
create_threads(10)
输出
executed 0. thread
executed 1. thread
executed 2. thread
executed 3. thread
executed 4. thread
executed 5. thread
executed 6. thread
executed 7. thread
executed 8. thread
executed 9. thread
对于那些喜欢根据输入列表使用 list comprehension 的人：
import threading
import time
maxthreads = 5
pool_sema = threading.Semaphore(value=maxthreads)
def task(i):
pool_sema.acquire()
try:
print("executed {}. thread".format(i))
time.sleep(2)
except Exception as e:
print("Error: problem with {0}. thread.\nMessage:{1}".format(i, e))
finally:
pool_sema.release()
def create_threads(number_of_threads):
try:
threads = [threading.Thread(target=task, args=(str(i))) for i in range(number_of_threads)]
[t.start() for t in threads]
except Exception as e:
print("Error: unable to start thread {}".format(e))
finally:
[t.join() for t in threads]
if __name__ == '__main__':
create_threads(10)
答案 7 :(得分：0)
使用 threading.activeCount() 方法限制最大线程数的简单方法
import threading, time
maxthreads = 10
def do_stuff(i):
print(i)
print("Total Active threads are {0}".format(threading.activeCount()))
time.sleep(20)
count = 0
while True:
if threading.activeCount() <= maxthreads:
worker = threading.Thread(target=do_stuff, args=(count,))
worker.start()
count += 1
答案 8 :(得分：-1)
要在主题创建时应用限制，请按照此示例（它确实有效）：
import threading
import time
def some_process(thread_num):
count = 0
while count < 5:
time.sleep(0.5)
count += 1
print "%s: %s" % (thread_num, time.ctime(time.time()))
print 'number of alive threads:{}'.format(threading.active_count())
def create_thread():
try:
for i in range(1, 555): # trying to spawn 555 threads.
thread = threading.Thread(target=some_process, args=(i,))
thread.start()
if threading.active_count() == 100: # set maximum threads.
thread.join()
print threading.active_count() # number of alive threads.
except Exception as e:
print "Error: unable to start thread {}".format(e)
if __name__ == '__main__':
create_thread()
或
另一种设置线程号检查器互斥/锁定的方法，如下例所示：
import threading
import time
def some_process(thread_num):
count = 0
while count < 5:
time.sleep(0.5)
count += 1
# print "%s: %s" % (thread_num, time.ctime(time.time()))
print 'number of alive threads:{}'.format(threading.active_count())
def create_thread2(number_of_desire_thread ):
try:
for i in range(1, 555):
thread = threading.Thread(target=some_process, args=(i,)).start()
while number_of_desire_thread <= threading.active_count():
'''mutex for avoiding to additional thread creation.'''
pass
print 'unlock'
print threading.active_count() # number of alive threads.
except Exception as e:
print "Error: unable to start thread {}".format(e)
if __name__ == '__main__':
create_thread2(100)