下面是运行任意命令返回其标准输出数据的Python代码，或在非零退出码上引发异常:
proc = subprocess.Popen(
cmd,
stderr=subprocess.STDOUT, # Merge stdout and stderr
stdout=subprocess.PIPE,
shell=True)
communication用于等待进程退出:
stdoutdata, stderrdata = proc.communicate()
子进程模块不支持超时——杀死运行超过X秒的进程的能力——因此，通信可能需要很长时间才能运行。
在Windows和Linux上运行的Python程序中实现超时的最简单方法是什么?
当前回答
我使用的解决方案是在shell命令前面加上timelimit。如果命令花费的时间太长，timelimit将停止它，Popen将有一个由timelimit设置的返回码。如果它是> 128，它意味着时间限制杀死进程。
参见python subprocess with timeout and large output (>64K)
其他回答
python 2.7
import time
import subprocess
def run_command(cmd, timeout=0):
start_time = time.time()
df = subprocess.Popen(cmd, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
while timeout and df.poll() == None:
if time.time()-start_time >= timeout:
df.kill()
return -1, ""
output = '\n'.join(df.communicate()).strip()
return df.returncode, output
我已经在Windows, Linux和Mac上成功地使用了killableprocess。如果你使用Cygwin Python，你将需要OSAF的killableprocess版本，否则本机Windows进程将不会被杀死。
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
我有一个问题，我想终止一个多线程子进程，如果它花费的时间超过给定的超时长度。我想在Popen()中设置一个超时，但它不起作用。然后，我意识到Popen().wait()等于call()，所以我有了在.wait(timeout=xxx)方法中设置超时的想法，这最终工作了。因此，我是这样解决的:
import os
import sys
import signal
import subprocess
from multiprocessing import Pool
cores_for_parallelization = 4
timeout_time = 15 # seconds
def main():
jobs = [...YOUR_JOB_LIST...]
with Pool(cores_for_parallelization) as p:
p.map(run_parallel_jobs, jobs)
def run_parallel_jobs(args):
# Define the arguments including the paths
initial_terminal_command = 'C:\\Python34\\python.exe' # Python executable
function_to_start = 'C:\\temp\\xyz.py' # The multithreading script
final_list = [initial_terminal_command, function_to_start]
final_list.extend(args)
# Start the subprocess and determine the process PID
subp = subprocess.Popen(final_list) # starts the process
pid = subp.pid
# Wait until the return code returns from the function by considering the timeout.
# If not, terminate the process.
try:
returncode = subp.wait(timeout=timeout_time) # should be zero if accomplished
except subprocess.TimeoutExpired:
# Distinguish between Linux and Windows and terminate the process if
# the timeout has been expired
if sys.platform == 'linux2':
os.kill(pid, signal.SIGTERM)
elif sys.platform == 'win32':
subp.terminate()
if __name__ == '__main__':
main()
对于python 2.6+，使用gevent
from gevent.subprocess import Popen, PIPE, STDOUT
def call_sys(cmd, timeout):
p= Popen(cmd, shell=True, stdout=PIPE)
output, _ = p.communicate(timeout=timeout)
assert p.returncode == 0, p. returncode
return output
call_sys('./t.sh', 2)
# t.sh example
sleep 5
echo done
exit 1
