Test if executable exists in Python?

I know this is an ancient question, but you can use distutils.spawn.find_executable. This has been documented since python 2.4 and has existed since python 1.6.

import distutils.spawndistutils.spawn.find_executable("notepad.exe")

Also, Python 3.3 now offers shutil.which().

Easiest way I can think of:

def which(program):    import os    def is_exe(fpath):        return os.path.isfile(fpath) and os.access(fpath, os.X_OK)    fpath, fname = os.path.split(program)    if fpath:        if is_exe(program):            return program    else:        for path in os.environ["PATH"].split(os.pathsep):            exe_file = os.path.join(path, program)            if is_exe(exe_file):                return exe_file    return None

Edit: Updated code sample to include logic for handling case where provided argument is already a full path to the executable, i.e. "which /bin/ls". This mimics the behavior of the UNIX 'which' command.

Edit: Updated to use os.path.isfile() instead of os.path.exists() per comments.

Edit: path.strip('"') seems like the wrong thing to do here. Neither Windows nor POSIX appear to encourage quoted PATH items.

Use shutil.which() from Python's wonderful standard library.Batteries included!