Python:如何并行运行python函数?

时间:2011-08-26 15:46:46

标签: python

我先研究过,找不到我的问题的答案。我试图在Python中并行运行多个函数。

我有这样的事情:

files.py

import common #common is a util class that handles all the IO stuff

dir1 = 'C:\folder1'
dir2 = 'C:\folder2'
filename = 'test.txt'
addFiles = [25, 5, 15, 35, 45, 25, 5, 15, 35, 45]

def func1():
   c = common.Common()
   for i in range(len(addFiles)):
       c.createFiles(addFiles[i], filename, dir1)
       c.getFiles(dir1)
       time.sleep(10)
       c.removeFiles(addFiles[i], dir1)
       c.getFiles(dir1)

def func2():
   c = common.Common()
   for i in range(len(addFiles)):
       c.createFiles(addFiles[i], filename, dir2)
       c.getFiles(dir2)
       time.sleep(10)
       c.removeFiles(addFiles[i], dir2)
       c.getFiles(dir2)

我想调用func1和func2并让它们同时运行。这些函数不会相互交互或在同一个对象上交互。现在我必须等待func1在func2启动之前完成。我如何做以下事情:

process.py

from files import func1, func2

runBothFunc(func1(), func2())

我希望能够非常接近同一时间创建两个目录,因为每分钟我都在计算正在创建的文件数量。如果目录不在那里,它会甩掉我的时间。

7 个答案:

答案 0 :(得分:114)

您可以使用threadingmultiprocessing

由于peculiarities of CPythonthreading不太可能实现真正的并行性。出于这个原因,multiprocessing通常是更好的选择。

这是一个完整的例子:

from multiprocessing import Process

def func1():
  print 'func1: starting'
  for i in xrange(10000000): pass
  print 'func1: finishing'

def func2():
  print 'func2: starting'
  for i in xrange(10000000): pass
  print 'func2: finishing'

if __name__ == '__main__':
  p1 = Process(target=func1)
  p1.start()
  p2 = Process(target=func2)
  p2.start()
  p1.join()
  p2.join()

启动/加入子进程的机制可以很容易地封装到runBothFunc中的函数中:

def runInParallel(*fns):
  proc = []
  for fn in fns:
    p = Process(target=fn)
    p.start()
    proc.append(p)
  for p in proc:
    p.join()

runInParallel(func1, func2)

答案 1 :(得分:7)

似乎您有一个函数需要调用两个不同的参数。可以结合使用concurrent.futuresmap和Python 3.2 +

来完成此操作
import time
from concurrent.futures import ThreadPoolExecutor, ProcessPoolExecutor

def sleep_secs(seconds):
  time.sleep(seconds)
  print(f'{seconds} has been processed')

secs_list = [2,4, 6, 8, 10, 12]

现在,如果您的操作受IO限制,则可以这样使用ThreadPoolExecutor

with ThreadPoolExecutor() as executor:
  results = executor.map(sleep_secs, secs_list)

请注意,此处如何使用map将函数map移至参数列表。

现在,如果您的功能受CPU限制,则可以使用ProcessPoolExecutor

with ProcessPoolExecutor() as executor:
  results = executor.map(sleep_secs, secs_list)

如果不确定,可以简单地尝试两者,看看哪一个可以给您更好的结果。

最后,如果您希望打印出结果,只需执行以下操作即可:

with ThreadPoolExecutor() as executor:
  results = executor.map(sleep_secs, secs_list)
  for result in results:
    print(result)

答案 2 :(得分:5)

这可以通过Ray优雅地完成,该系统使您可以轻松地并行化和分发Python代码。

要并行化示例,您需要使用@ray.remote装饰器定义函数,然后使用.remote调用它们。

import ray

ray.init()

dir1 = 'C:\\folder1'
dir2 = 'C:\\folder2'
filename = 'test.txt'
addFiles = [25, 5, 15, 35, 45, 25, 5, 15, 35, 45]

# Define the functions. 
# You need to pass every global variable used by the function as an argument.
# This is needed because each remote function runs in a different process,
# and thus it does not have access to the global variables defined in 
# the current process.
@ray.remote
def func1(filename, addFiles, dir):
    # func1() code here...

@ray.remote
def func2(filename, addFiles, dir):
    # func2() code here...

# Start two tasks in the background and wait for them to finish.
ray.get([func1.remote(filename, addFiles, dir1), func2.remote(filename, addFiles, dir2)]) 

如果将相同的参数传递给两个函数并且参数较大,则更有效的方法是使用ray.put()。这样可以避免将大参数序列化两次并为其创建两个内存副本:

largeData_id = ray.put(largeData)

ray.get([func1(largeData_id), func2(largeData_id)])

如果func1()func2()返回结果,则需要按以下方式重写代码:

ret_id1 = func1.remote(filename, addFiles, dir1)
ret_id2 = func1.remote(filename, addFiles, dir2)
ret1, ret2 = ray.get([ret_id1, ret_id2])

multiprocessing模块相比,使用Ray有许多优点。特别是,相同的代码将在单台计算机以及多台计算机上运行。有关Ray的更多优点,请参见this related post

答案 3 :(得分:4)

2021 年最简单的方法是使用 asyncio:

import asyncio, time

async def say_after(delay, what):
    await asyncio.sleep(delay)
    print(what)

async def main():

    task1 = asyncio.create_task(
        say_after(4, 'hello'))

    task2 = asyncio.create_task(
        say_after(3, 'world'))

    print(f"started at {time.strftime('%X')}")

    # Wait until both tasks are completed (should take
    # around 2 seconds.)
    await task1
    await task2

    print(f"finished at {time.strftime('%X')}")


asyncio.run(main())

参考:

[1] https://docs.python.org/3/library/asyncio-task.html

答案 4 :(得分:3)

没有办法保证两个函数会彼此同步执行,这似乎是你想要做的事情。

您可以做的最好的事情是将该功能分成几个步骤,然后等待使用Process.join完成关键同步点,如@ aix的答案提及。

这比time.sleep(10)更好,因为您无法保证准确的时间安排。在明确等待的情况下,你说这些函数必须在移动到下一步之前执行该步骤,而不是假设它将在10ms内完成,而这根据机器上的其他内容无法保证。

答案 5 :(得分:3)

如果您是Windows用户并使用python 3,那么这篇文章将帮助您在python中进行并行编程。当您运行通常的多处理库的池编程时,您将收到有关程序中主函数的错误。这是因为windows没有fork()功能。以下帖子正在解决上述问题。

http://python.6.x6.nabble.com/Multiprocessing-Pool-woes-td5047050.html

因为我使用的是python 3,所以我改变了这个程序:

from types import FunctionType
import marshal

def _applicable(*args, **kwargs):
  name = kwargs['__pw_name']
  code = marshal.loads(kwargs['__pw_code'])
  gbls = globals() #gbls = marshal.loads(kwargs['__pw_gbls'])
  defs = marshal.loads(kwargs['__pw_defs'])
  clsr = marshal.loads(kwargs['__pw_clsr'])
  fdct = marshal.loads(kwargs['__pw_fdct'])
  func = FunctionType(code, gbls, name, defs, clsr)
  func.fdct = fdct
  del kwargs['__pw_name']
  del kwargs['__pw_code']
  del kwargs['__pw_defs']
  del kwargs['__pw_clsr']
  del kwargs['__pw_fdct']
  return func(*args, **kwargs)

def make_applicable(f, *args, **kwargs):
  if not isinstance(f, FunctionType): raise ValueError('argument must be a function')
  kwargs['__pw_name'] = f.__name__  # edited
  kwargs['__pw_code'] = marshal.dumps(f.__code__)   # edited
  kwargs['__pw_defs'] = marshal.dumps(f.__defaults__)  # edited
  kwargs['__pw_clsr'] = marshal.dumps(f.__closure__)  # edited
  kwargs['__pw_fdct'] = marshal.dumps(f.__dict__)   # edited
  return _applicable, args, kwargs

def _mappable(x):
  x,name,code,defs,clsr,fdct = x
  code = marshal.loads(code)
  gbls = globals() #gbls = marshal.loads(gbls)
  defs = marshal.loads(defs)
  clsr = marshal.loads(clsr)
  fdct = marshal.loads(fdct)
  func = FunctionType(code, gbls, name, defs, clsr)
  func.fdct = fdct
  return func(x)

def make_mappable(f, iterable):
  if not isinstance(f, FunctionType): raise ValueError('argument must be a function')
  name = f.__name__    # edited
  code = marshal.dumps(f.__code__)   # edited
  defs = marshal.dumps(f.__defaults__)  # edited
  clsr = marshal.dumps(f.__closure__)  # edited
  fdct = marshal.dumps(f.__dict__)  # edited
  return _mappable, ((i,name,code,defs,clsr,fdct) for i in iterable)

在此功能之后,上面的问题代码也改变了一点:

from multiprocessing import Pool
from poolable import make_applicable, make_mappable

def cube(x):
  return x**3

if __name__ == "__main__":
  pool    = Pool(processes=2)
  results = [pool.apply_async(*make_applicable(cube,x)) for x in range(1,7)]
  print([result.get(timeout=10) for result in results])

我得到了输出:

[1, 8, 27, 64, 125, 216]

我认为这篇文章可能对某些Windows用户有用。

答案 6 :(得分:1)

如果您的函数主要用于完成 I / O工作(并且CPU工作量较少)并且您使用的是Python 3.2+,则可以使用ThreadPoolExecutor

from concurrent.futures import ThreadPoolExecutor

def run_io_tasks_in_parallel(tasks):
    with ThreadPoolExecutor() as executor:
        running_tasks = [executor.submit(task) for task in tasks]
        for running_task in running_tasks:
            running_task.result()

run_io_tasks_in_parallel([
    lambda: print('IO task 1 running!'),
    lambda: print('IO task 2 running!'),
])

如果您的功能主要用于完成 CPU工作(并且I / O工作较少)并且您具有Python 2.6+,则可以使用multiprocessing模块:

from multiprocessing import Process

def run_cpu_tasks_in_parallel(tasks):
    running_tasks = [Process(target=task) for task in tasks]
    for running_task in running_tasks:
        running_task.start()
    for running_task in running_tasks:
        running_task.join()

run_cpu_tasks_in_parallel([
    lambda: print('CPU task 1 running!'),
    lambda: print('CPU task 2 running!'),
])