Question

如何在没有主动等待（轮询）的情况下在Windows上等待Python中的多个子进程？像几乎这样的东西对我有用：

proc1 = subprocess.Popen(['python','mytest.py'])
proc2 = subprocess.Popen(['python','mytest.py'])    
proc1.wait()
print "1 finished"
proc2.wait()
print "2 finished"

问题是当proc2在proc1之前完成时，父进程仍会等待proc1。在Unix上，可以在循环中使用waitpid(0)来获取子进程的返回代码 - 如何在Windows上用Python实现类似的东西？

Answer 1

这看起来有点矫枉过正，但是，在这里：

import Queue, thread, subprocess

results= Queue.Queue()
def process_waiter(popen, description, que):
    try: popen.wait()
    finally: que.put( (description, popen.returncode) )
process_count= 0

proc1= subprocess.Popen( ['python', 'mytest.py'] )
thread.start_new_thread(process_waiter,
    (proc1, "1 finished", results))
process_count+= 1

proc2= subprocess.Popen( ['python', 'mytest.py'] )
thread.start_new_thread(process_waiter,
    (proc2, "2 finished", results))
process_count+= 1

# etc

while process_count > 0:
    description, rc= results.get()
    print "job", description, "ended with rc =", rc
    process_count-= 1

Answer 2

Twisted的asynchronous process-spawning API适用于Windows。实际上有几种不同的实现，其中许多并不是很好，但您可以在不改变代码的情况下在它们之间切换。

Answer 3

基于zseil的答案，您可以通过混合使用子进程和win32 API来实现此目的。我使用直接ctypes，因为我的Python没有安装win32api。我现在只是从MSYS中生成sleep.exe作为一个例子，但显然你可以生成任何你喜欢的进程。我使用OpenProcess（）从进程'PID获取一个HANDLE，然后使用WaitForMultipleObjects等待任何进程完成。

import ctypes, subprocess
from random import randint
SYNCHRONIZE=0x00100000
INFINITE = -1
numprocs = 5
handles = {}

for i in xrange(numprocs):
    sleeptime = randint(5,10)
    p = subprocess.Popen([r"c:\msys\1.0\bin\sleep.exe", str(sleeptime)], stdin=subprocess.PIPE, stdout=subprocess.PIPE, stderr=subprocess.PIPE, shell=False)
    h = ctypes.windll.kernel32.OpenProcess(SYNCHRONIZE, False, p.pid)
    handles[h] = p.pid
    print "Spawned Process %d" % p.pid

while len(handles) > 0:
    print "Waiting for %d children..." % len(handles)
    arrtype = ctypes.c_long * len(handles)
    handle_array = arrtype(*handles.keys())
    ret = ctypes.windll.kernel32.WaitForMultipleObjects(len(handle_array), handle_array, False, INFINITE)
    h = handle_array[ret]
    ctypes.windll.kernel32.CloseHandle(h)
    print "Process %d done" % handles[h]
    del handles[h]
print "All done!"

Answer 4

Windows上的Twisted将执行主动等待。如果您不想使用线程，则必须使用win32 API来避免轮询。像这样：

import win32process
import win32event

# Note: CreateProcess() args are somewhat cryptic, look them up on MSDN
proc1, thread1, pid1, tid1 = win32process.CreateProcess(...)
proc2, thread2, pid2, tid2 = win32process.CreateProcess(...)
thread1.close()
thread2.close()

processes = {proc1: "proc1", proc2: "proc2"}

while processes:
    handles = processes.keys()
    # Note: WaitForMultipleObjects() supports at most 64 processes at a time
    index = win32event.WaitForMultipleObjects(handles, False, win32event.INFINITE)
    finished = handles[index]
    exitcode = win32process.GetExitCodeProcess(finished)
    procname = processes.pop(finished)
    finished.close()
    print "Subprocess %s finished with exit code %d" % (procname, exitcode)

Answer 5

您可以使用psutil：

>>> import subprocess
>>> import psutil
>>> 
>>> proc1 = subprocess.Popen(['python','mytest.py'])
>>> proc2 = subprocess.Popen(['python','mytest.py'])    
>>> ls = [psutil.Process(proc1.pid), psutil.Process(proc2.pid)]
>>>
>>> gone, alive = psutil.wait_procs(ls, timeout=3)

'消失'和'活着'是指示哪些进程消失以及哪些进程仍然存活的列表。

您可以选择指定每次观看的某个进程终止时调用的回调：

>>> def on_terminate(proc):
...     print "%s terminated" % proc
...
>>> gone, alive = psutil.wait_procs(ls, timeout=3, callback=on_terminate)

Answer 6

您可以使用psutil

import psutil

with psutil.Popen(["python", "mytest.py"]) as proc1, psutil.Popen(
    ["python", "mytest.py"]
) as proc2:
    gone, alive = psutil.wait_procs([proc1, proc2], timeout=3)

'gone'和'alive'是指示哪些进程已消失而哪些进程仍在运行的列表。

您可以选择指定一个回调，该回调在每次被监视进程之一终止时调用：

def on_terminate(proc):
    print "%s terminated" % proc

gone, alive = psutil.wait_procs(ls, timeout=3, callback=on_terminate)

Windows上的Python - 如何等待多个子进程？

6 个答案: