How to check if there exists a process with a given pid in Python?

Question

Is there a way to check to see if a pid corresponds to a valid process? I'm getting a pid from a different source other than from os.getpid() and I need to check to see if a process with that pid doesn't exist on the machine.

I need it to be available in Unix and Windows. I'm also checking to see if the PID is NOT in use.

Answer 1

Sending signal 0 to a pid will raise an OSError exception if the pid is not running, and do nothing otherwise.

import os

def check_pid(pid):        
    """ Check For the existence of a unix pid. """
    try:
        os.kill(pid, 0)
    except OSError:
        return False
    else:
        return True

Answer 2

Have a look at the psutil module:

psutil (python system and process utilities) is a cross-platform library for retrieving information on running processes and system utilization (CPU, memory, disks, network) in Python. [...] It currently supports Linux, Windows, OSX, FreeBSD and Sun Solaris, both 32-bit and 64-bit architectures, with Python versions from 2.6 to 3.4 (users of Python 2.4 and 2.5 may use 2.1.3 version). PyPy is also known to work.

It has a function called pid_exists() that you can use to check whether a process with the given pid exists.

Here's an example:

import psutil
pid = 12345
if psutil.pid_exists(pid):
    print "a process with pid %d exists" % pid
else:
    print "a process with pid %d does not exist" % pid

For reference:

• https://pypi.python.org/pypi/psutil
• https://github.com/giampaolo/psutil
• http://pythonhosted.org/psutil/#psutil.pid_exists

Answer 3

mluebke code is not 100% correct; kill() can also raise EPERM (access denied) in which case that obviously means a process exists. This is supposed to work:

(edited as per Jason R. Coombs comments)

import errno
import os
import sys

def pid_exists(pid):
    """Check whether pid exists in the current process table.
    UNIX only.
    """
    if pid < 0:
        return False
    if pid == 0:
        # According to "man 2 kill" PID 0 refers to every process
        # in the process group of the calling process.
        # On certain systems 0 is a valid PID but we have no way
        # to know that in a portable fashion.
        raise ValueError('invalid PID 0')
    try:
        os.kill(pid, 0)
    except OSError as err:
        if err.errno == errno.ESRCH:
            # ESRCH == No such process
            return False
        elif err.errno == errno.EPERM:
            # EPERM clearly means there's a process to deny access to
            return True
        else:
            # According to "man 2 kill" possible error values are
            # (EINVAL, EPERM, ESRCH)
            raise
    else:
        return True

You can't do this on Windows unless you use pywin32, ctypes or a C extension module. If you're OK with depending from an external lib you can use psutil:

>>> import psutil
>>> psutil.pid_exists(2353)
True

Answer 4

The answers involving sending 'signal 0' to the process will work only if the process in question is owned by the user running the test. Otherwise you will get an OSError due to permissions, even if the pid exists in the system.

In order to bypass this limitation you can check if /proc/<pid> exists:

import os

def is_running(pid):
    if os.path.isdir('/proc/{}'.format(pid)):
        return True
    return False

This applies to linux based systems only, obviously.

Answer 5

Look here for windows-specific way of getting full list of running processes with their IDs. It would be something like

from win32com.client import GetObject
def get_proclist():
    WMI = GetObject('winmgmts:')
    processes = WMI.InstancesOf('Win32_Process')
    return [process.Properties_('ProcessID').Value for process in processes]

You can then verify pid you get against this list. I have no idea about performance cost, so you'd better check this if you're going to do pid verification often.

For *NIx, just use mluebke's solution.

Answer 6

In Python 3.3+, you could use exception names instead of errno constants. Posix version:

import os

def pid_exists(pid): 
    if pid < 0: return False #NOTE: pid == 0 returns True
    try:
        os.kill(pid, 0) 
    except ProcessLookupError: # errno.ESRCH
        return False # No such process
    except PermissionError: # errno.EPERM
        return True # Operation not permitted (i.e., process exists)
    else:
        return True # no error, we can send a signal to the process

Answer 7

Building upon ntrrgc's I've beefed up the windows version so it checks the process exit code and checks for permissions:

def pid_exists(pid):
    """Check whether pid exists in the current process table."""
    if os.name == 'posix':
        import errno
        if pid < 0:
            return False
        try:
            os.kill(pid, 0)
        except OSError as e:
            return e.errno == errno.EPERM
        else:
            return True
    else:
        import ctypes
        kernel32 = ctypes.windll.kernel32
        HANDLE = ctypes.c_void_p
        DWORD = ctypes.c_ulong
        LPDWORD = ctypes.POINTER(DWORD)
        class ExitCodeProcess(ctypes.Structure):
            _fields_ = [ ('hProcess', HANDLE),
                ('lpExitCode', LPDWORD)]

        SYNCHRONIZE = 0x100000
        process = kernel32.OpenProcess(SYNCHRONIZE, 0, pid)
        if not process:
            return False

        ec = ExitCodeProcess()
        out = kernel32.GetExitCodeProcess(process, ctypes.byref(ec))
        if not out:
            err = kernel32.GetLastError()
            if kernel32.GetLastError() == 5:
                # Access is denied.
                logging.warning("Access is denied to get pid info.")
            kernel32.CloseHandle(process)
            return False
        elif bool(ec.lpExitCode):
            # print ec.lpExitCode.contents
            # There is an exist code, it quit
            kernel32.CloseHandle(process)
            return False
        # No exit code, it's running.
        kernel32.CloseHandle(process)
        return True

Answer 8

Combining Giampaolo Rodolà's answer for POSIX and mine for Windows I got this:

import os
if os.name == 'posix':
    def pid_exists(pid):
        """Check whether pid exists in the current process table."""
        import errno
        if pid < 0:
            return False
        try:
            os.kill(pid, 0)
        except OSError as e:
            return e.errno == errno.EPERM
        else:
            return True
else:
    def pid_exists(pid):
        import ctypes
        kernel32 = ctypes.windll.kernel32
        SYNCHRONIZE = 0x100000

        process = kernel32.OpenProcess(SYNCHRONIZE, 0, pid)
        if process != 0:
            kernel32.CloseHandle(process)
            return True
        else:
            return False

Answer 9

In Windows, you can do it in this way:

import ctypes
PROCESS_QUERY_INFROMATION = 0x1000
def checkPid(pid):
    processHandle = ctypes.windll.kernel32.OpenProcess(PROCESS_QUERY_INFROMATION, 0,pid)
    if processHandle == 0:
        return False
    else:
        ctypes.windll.kernel32.CloseHandle(processHandle)
    return True

First of all, in this code you try to get a handle for process with pid given. If the handle is valid, then close the handle for process and return True; otherwise, you return False. Documentation for OpenProcess: https://msdn.microsoft.com/en-us/library/windows/desktop/ms684320%28v=vs.85%29.aspx

Answer 10

This will work for Linux, for example if you want to check if banshee is running... (banshee is a music player)

import subprocess

def running_process(process):
    "check if process is running. < process > is the name of the process."

    proc = subprocess.Popen(["if pgrep " + process + " >/dev/null 2>&1; then echo 'True'; else echo 'False'; fi"], stdout=subprocess.PIPE, shell=True)

    (Process_Existance, err) = proc.communicate()
    return Process_Existance

# use the function
print running_process("banshee")

Answer 11

I'd say use the PID for whatever purpose you're obtaining it and handle the errors gracefully. Otherwise, it's a classic race (the PID may be valid when you check it's valid, but go away an instant later)