我想创建一个运行多个轻线程的程序，但将自身限制为恒定的、预定义数量的并发运行任务，如下所示（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()
实现这一点的最安全/最快的方法是什么？
听起来你想用八名工人实现生产者/消费者模式。Python 有一个Queue用于此目的的类，它是线程安全的。
每个工作人员都应该调用get()队列来检索任务。如果没有可用的任务，此调用将阻塞，导致工作人员空闲，直到有一个可用。然后工作人员应该执行任务并最终调用task_done()队列。
您可以通过调用队列将任务放入put()队列中。
在主线程中，您可以调用join()队列以等待所有待处理的任务完成。
这种方法的好处是您不会创建和销毁线程，这很昂贵。工作线程将连续运行，但当队列中没有任务时将进入睡眠状态，使用零 CPU 时间。
（链接的文档页面有一个这种模式的例子。）
信号量是一种变量或抽象数据类型，用于控制并发系统（如多道程序操作系统）中多个进程对公共资源的访问；这可以帮助你。
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>
通过这种方式，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“maximumNumberOfThreads”可用于定义线程最大值的上限。
我遇到了同样的问题，并花了几天（准确地说是 2 天）使用队列找到正确的解决方案。我在 ThreadPoolExecutor 路径上浪费了一天，因为没有办法限制事物启动的线程数！我给它提供了一个包含 5000 个要复制的文件的列表，一旦它同时运行了大约 1500 个并发文件副本，代码就会无响应。ThreadPoolExecutor 上的 max_workers 参数仅控制有多少工作线程正在启动线程，而不是有多少线程启动。
好的，无论如何，这是一个非常简单的使用队列的例子：
import threading, time, random
from queue import Queue
jobs = Queue()
def do_stuff(q):
while not q.empty():
value = q.get()
time.sleep(random.randint(1, 10))
print(value)
q.task_done()
for i in range(10):
jobs.put(i)
for i in range(3):
worker = threading.Thread(target=do_stuff, args=(jobs,))
worker.start()
print("waiting for queue to complete", jobs.qsize(), "tasks")
jobs.join()
print("all done")
multiprocessing.dummy.Pool使用, 或concurrent.futures.ThreadPoolExecutor（或者，如果使用 Python 2.x，则使用 backport ）将其实现为线程池或执行器会容易得多futures。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更改get_task为推模型get_tasks，例如一次产生一个任务，则更简单：
with concurrent.futures.ThreadPoolExecutor(8) as executor:
for arg in get_tasks():
executor.submit(f, arg)
当您用完任务（例如，get_task引发异常或get_tasks干涸）时，这将自动告诉执行程序在排空队列后停止，等待它停止并清理所有内容。
concurrent.futures.ThreadPoolExecutor.map
concurrent.futures.ThreadPoolExecutor在https://stackoverflow.com/a/19370282/895245map中提到过，这里有一个通常是最方便的方法的示例。
.map()是 的并行版本map()：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
给出：
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
给出：
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
所以请注意它是如何在异常时立即停止的。
Queue错误处理示例
Queue在https://stackoverflow.com/a/19369877/895245中提到过，但这里有一个完整的例子。
设计目标：
concurrent.futures.ThreadPoolExecutor是我见过的 stdlib 中当前可用的最好的接口。但是我找不到如何执行以下所有操作：
因为：
.map(): 一次读取所有输入并且func只能接受参数.submit()：.shutdown()执行直到所有期货完成，并且最大当前工作项没有阻塞.submit()。那么如何避免.cancel()在第一次失败后对所有期货产生丑陋的循环呢？事不宜迟，这是我的实现。测试用例在脚本末尾的下面__name__ == '__main__'：
线程池.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 中测试。
我见过最常见的写法是：
threads = [threading.Thread(target=f) for _ in range(8)]
for thread in threads:
thread.start()
...
for thread in threads:
thread.join()
如果您想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请求新工作，请考虑围绕队列构建的解决方案，例如“如何等待 Python 中只有第一个线程完成”。
使用 threading.activeCount() 方法限制最大线程的简单和最简单的方法
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
它可以很容易地使用ThreadPoolExecutor. 使用max_workers参数更改限制。
from concurrent.futures import ThreadPoolExecutor
import time
pool = ThreadPoolExecutor(max_workers=10)
def thread(num):
print(num)
time.sleep(3)
for n in range(0, 1000):
pool.submit(thread, n)
pool.shutdown(wait=True)
这可以通过Semaphore Object来完成。信号量管理一个内部计数器，每次调用递减，每次acquire()调用递增release()。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()。
一个简短的示例显示了最多 5 个并行线程，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
对于那些喜欢基于输入列表使用列表推导的人：
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)
对于线程创建的应用限制，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()
或者：
另一种设置线程号检查器互斥/锁的方法，例如以下示例：
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)