Test if executable exists in Python?

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.

Just remember to specify the file extension on windows. Otherwise, you have to write a much complicated is_exe for windows using PATHEXT environment variable. You may just want to use FindPath.

OTOH, why are you even bothering to search for the executable? The operating system will do it for you as part of popen call & will raise an exception if the executable is not found. All you need to do is catch the correct exception for given OS. Note that on Windows, subprocess.Popen(exe, shell=True) will fail silently if exe is not found.

Incorporating PATHEXT into the above implementation of which (in Jay's answer):

def which(program):
    def is_exe(fpath):
        return os.path.exists(fpath) and os.access(fpath, os.X_OK) and os.path.isfile(fpath)

    def ext_candidates(fpath):
        yield fpath
        for ext in os.environ.get("PATHEXT", "").split(os.pathsep):
            yield fpath + ext

    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)
            for candidate in ext_candidates(exe_file):
                if is_exe(candidate):
                    return candidate

    return None

See os.path module for some useful functions on pathnames. To check if an existing file is executable, use os.access(path, mode), with the os.X_OK mode.

os.X_OK

Value to include in the mode parameter of access() to determine if path can be executed.

EDIT: The suggested which() implementations are missing one clue - using os.path.join() to build full file names.

So basically you want to find a file in mounted filesystem (not necessarily in PATH directories only) and check if it is executable. This translates to following plan:

• enumerate all files in locally mounted filesystems
• match results with name pattern
• for each file found check if it is executable

I'd say, doing this in a portable way will require lots of computing power and time. Is it really what you need?

There is a which.py script in a standard Python distribution (e.g. on Windows '\PythonXX\Tools\Scripts\which.py').

EDIT: which.py depends on ls therefore it is not cross-platform.