我添加了从jcollado线程到我的Python模块easyprocess的解决方案。

安装:

pip install easyprocess

例子:

from easyprocess import Proc

# shell is not supported!
stdout=Proc('ping localhost').call(timeout=1.5).stdout
print stdout

This solution kills the process tree in case of shell=True, passes parameters to the process (or not), has a timeout and gets the stdout, stderr and process output of the call back (it uses psutil for the kill_proc_tree). This was based on several solutions posted in SO including jcollado's. Posting in response to comments by Anson and jradice in jcollado's answer. Tested in Windows Srvr 2012 and Ubuntu 14.04. Please note that for Ubuntu you need to change the parent.children(...) call to parent.get_children(...).

def kill_proc_tree(pid, including_parent=True):
  parent = psutil.Process(pid)
  children = parent.children(recursive=True)
  for child in children:
    child.kill()
  psutil.wait_procs(children, timeout=5)
  if including_parent:
    parent.kill()
    parent.wait(5)

def run_with_timeout(cmd, current_dir, cmd_parms, timeout):
  def target():
    process = subprocess.Popen(cmd, cwd=current_dir, shell=True, stdout=subprocess.PIPE, stdin=subprocess.PIPE, stderr=subprocess.PIPE)

    # wait for the process to terminate
    if (cmd_parms == ""):
      out, err = process.communicate()
    else:
      out, err = process.communicate(cmd_parms)
    errcode = process.returncode

  thread = Thread(target=target)
  thread.start()

  thread.join(timeout)
  if thread.is_alive():
    me = os.getpid()
    kill_proc_tree(me, including_parent=False)
    thread.join()

Jcollado的答案可以使用线程来简化。定时器类:

import shlex
from subprocess import Popen, PIPE
from threading import Timer

def run(cmd, timeout_sec):
    proc = Popen(shlex.split(cmd), stdout=PIPE, stderr=PIPE)
    timer = Timer(timeout_sec, proc.kill)
    try:
        timer.start()
        stdout, stderr = proc.communicate()
    finally:
        timer.cancel()

# Examples: both take 1 second
run("sleep 1", 5)  # process ends normally at 1 second
run("sleep 5", 1)  # timeout happens at 1 second

有一个想法是子类化Popen类并用一些简单的方法装饰器来扩展它。我们叫它ExpirablePopen吧。

from logging import error
from subprocess import Popen
from threading import Event
from threading import Thread


class ExpirablePopen(Popen):

    def __init__(self, *args, **kwargs):
        self.timeout = kwargs.pop('timeout', 0)
        self.timer = None
        self.done = Event()

        Popen.__init__(self, *args, **kwargs)

    def __tkill(self):
        timeout = self.timeout
        if not self.done.wait(timeout):
            error('Terminating process {} by timeout of {} secs.'.format(self.pid, timeout))
            self.kill()

    def expirable(func):
        def wrapper(self, *args, **kwargs):
            # zero timeout means call of parent method
            if self.timeout == 0:
                return func(self, *args, **kwargs)

            # if timer is None, need to start it
            if self.timer is None:
                self.timer = thr = Thread(target=self.__tkill)
                thr.daemon = True
                thr.start()

            result = func(self, *args, **kwargs)
            self.done.set()

            return result
        return wrapper

    wait = expirable(Popen.wait)
    communicate = expirable(Popen.communicate)


if __name__ == '__main__':
    from subprocess import PIPE

    print ExpirablePopen('ssh -T git@bitbucket.org', stdout=PIPE, timeout=1).communicate()

没想到居然没人提到超时

Timeout 5 ping -c 3 somehost

显然，这并不适用于每个用例，但如果您处理的是一个简单的脚本，那么这是很难克服的。

mac用户也可以通过homebrew在coreutils中使用gtimeout。

如果你用的是Unix，

import signal
  ...
class Alarm(Exception):
    pass

def alarm_handler(signum, frame):
    raise Alarm

signal.signal(signal.SIGALRM, alarm_handler)
signal.alarm(5*60)  # 5 minutes
try:
    stdoutdata, stderrdata = proc.communicate()
    signal.alarm(0)  # reset the alarm
except Alarm:
    print "Oops, taking too long!"
    # whatever else