Test if executable exists in Python?

Question

In Python, is there a portable and simple way to test if an executable program exists?

By simple I mean something like the which command which would be just perfect. I don't want to search PATH manually or something involving trying to execute it with Popen & al and see if it fails (that's what I'm doing now, but imagine it's launchmissiles)

Answer 1

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?

Answer 2

I found something in StackOverflow that solved the problem for me. This works provided the executable has an option (like --help or --version) that outputs something and returns an exit status of zero. See Suppress output in Python calls to executables - the "result" at the end of the code snippet in this answer will be zero if the executable is in path, else it is most likely to be 1.

Answer 3

An important question is "Why do you need to test if executable exist?" Maybe you don't? ;-)

Recently I needed this functionality to launch viewer for PNG file. I wanted to iterate over some predefined viewers and run the first that exists. Fortunately, I came across os.startfile. It's much better! Simple, portable and uses the default viewer on the system:

>>> os.startfile('yourfile.png')

Update: I was wrong about os.startfile being portable... It's Windows only. On Mac you have to run open command. And xdg_open on Unix. There's a Python issue on adding Mac and Unix support for os.startfile.

Answer 4

It would seem the obvious choice is "which", parsing the results via popen, but you could simulate it otherwise using the os class. In pseudopython, it would look like this:

for each element r in path:
    for each file f in directory p:
        if f is executable:
           return True

Answer 5

Python 3.3 now offers shutil.which().

Answer 6

For *nix platforms (Linux and OS X)

This seems to be working for me:

Edited to work on Linux, thanks to Mestreion

def cmd_exists(cmd):
    return subprocess.call("type " + cmd, shell=True, 
        stdout=subprocess.PIPE, stderr=subprocess.PIPE) == 0

What we're doing here is using the builtin command type and checking the exit code. If there's no such command, type will exit with 1 (or a non-zero status code anyway).

The bit about stdout and stderr is just to silence the output of the type command, since we're only interested in the exit status code.

Example usage:

>>> cmd_exists("jsmin")
True
>>> cmd_exists("cssmin")
False
>>> cmd_exists("ls")
True
>>> cmd_exists("dir")
False
>>> cmd_exists("node")
True
>>> cmd_exists("steam")
False