Check if a program exists from a python script [duplicate]

Answer 1

The easiest way is to try to run the program with the desired parameters, and handle the exception if it doesn't exist:

try:
    subprocess.call(["wget", "your", "parameters", "here"])
except OSError as e:
    if e.errno == errno.ENOENT:
        # handle file not found error.
    else:
        # Something else went wrong while trying to run `wget`
        raise

This is a common pattern in Python: EAFP

Answer 2

import subprocess
import os

def is_tool(name):
    try:
        devnull = open(os.devnull)
        subprocess.Popen([name], stdout=devnull, stderr=devnull).communicate()
    except OSError as e:
        if e.errno == os.errno.ENOENT:
            return False
    return True

Answer 3

A slight modification to @SvenMarnach's code that addresses the issue of printing to the standard output stream. If you use the subprocess.check_output() function rather than subprocess.call() then you can handle the string that is normally printed to standard out in your code and still catch exceptions and the exit status code.

If you want to suppress the standard output stream in the terminal, don’t print the std out string that is returned from check_output:

import subprocess
import os
try:
    stdout_string = subprocess.check_output(["wget", "--help"], stderr=subprocess.STDOUT)
    # print(stdout_string)
except subprocess.CalledProcessError as cpe:
    print(cpe.returncode)
    print(cpe.output)
except OSError as e:
    if e.errno == os.errno.ENOENT:
        print(e)
    else:
        # Something else went wrong while trying to run `wget`
        print(e)

The non-zero exit status code and output string are raised in the CalledProcessError as subprocess.CalledProcessError.returncode and subprocess.CalledProcessError.output so you can do whatever you'd like with them.

If you want to print the executable's standard output to the terminal, print the string that is returned:

import subprocess
import os
try:
    stdout_string = subprocess.check_output(["wget", "--help"], stderr=subprocess.STDOUT)
    print(stdout_string)
except subprocess.CalledProcessError as cpe:
    print(cpe.returncode)
    print(cpe.output)
except OSError as e:
    if e.errno == os.errno.ENOENT:
        print(e)
    else:
        # Something else went wrong while trying to run `wget`
        print(e)

print() adds an extra newline to the string. If you want to eliminate that (and write std error to the std err stream instead of the std out stream as shown with the print() statements above), use sys.stdout.write(string) and sys.stderr.write(string) instead of print():

import subprocess
import os
import sys
try:
    stdout_string = subprocess.check_output(["bogus"], stderr=subprocess.STDOUT)
    sys.stdout.write(stdout_string)
except subprocess.CalledProcessError as cpe:
    sys.stderr.write(cpe.returncode)
    sys.stderr.write(cpe.output)
except OSError as e:
    if e.errno == os.errno.ENOENT:
        sys.stderr.write(e.strerror)
    else:
        # Something else went wrong while trying to run `wget`
        sys.stderr.write(e.strerror)

Answer 4

I would probably shell out to which wget or which curl and check that the result ends in the name of the program you are using. The magic of unix :)

Actually, all you need to do is check the return code of which. So... using our trusty subprocess module:

import subprocess

rc = subprocess.call(['which', 'wget'])
if rc == 0:
    print 'wget installed!'
else:
    print 'wget missing in path!'

Note that I tested this on windows with cygwin... If you want to figure out how to implement which in pure python, i suggest you check here: http://pypi.python.org/pypi/pycoreutils (oh dear - it seems they don't supply which. Time for a friendly nudge?)

UPDATE: On Windows, you can use where instead of which for a similar effect.

Answer 5

import os
import subprocess


def is_tool(prog):
    for dir in os.environ['PATH'].split(os.pathsep):
        if os.path.exists(os.path.join(dir, prog)):
            try:
                subprocess.call([os.path.join(dir, prog)],
                                stdout=subprocess.PIPE,
                                stderr=subprocess.STDOUT)
            except OSError, e:
                return False
            return True
    return False

Answer 6

I'd change @sorin's answer as follows, the reason is it would check the name of the program without passing the absolute path of the program

from subprocess import Popen, PIPE

def check_program_exists(name):
    p = Popen(['/usr/bin/which', name], stdout=PIPE, stderr=PIPE)
    p.communicate()
    return p.returncode == 0