在Python 3.7.8中超时测试后捕获的输出示例:

try:
    return subprocess.run(command, shell=True, capture_output=True, timeout=20, cwd=cwd, universal_newlines=True)
except subprocess.TimeoutExpired as e:
    print(e.output.decode(encoding="utf-8", errors="ignore"))
    assert False;

异常子进程。TimeoutExpired有输出和其他成员:

cmd -用于生成子进程的命令。 timeout -超时时间，单位为秒。 output -子进程的输出，如果它是由run()或 check_output()。否则,没有。 stdout -输出的别名，用于与stderr对称。 stderr -子进程的stderr输出，如果它被捕获 run()。否则,没有。

更多信息:https://docs.python.org/3/library/subprocess.html#subprocess.TimeoutExpired

从Python 3.5开始，有了一个新的subprocess.run通用命令(意味着取代check_call, check_output…)，它也有timeout=参数。

subprocess.run(args， *， stdin=None, input=None, stdout=None, stderr=None, shell=False, cwd=None, timeout=None, check=False, encoding=None, errors=None) 执行args描述的命令。等待命令完成，然后返回一个CompletedProcess实例。

它引发一个子流程。超时超时时出现TimeoutExpired异常。

如果你正在使用python2，请尝试一下

import subprocess32

try:
    output = subprocess32.check_output(command, shell=True, timeout=3)
except subprocess32.TimeoutExpired as e:
    print e

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

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()

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