Python error when executing shell script

Question

I'm executing a line of shell script using python. It isn't working with which to test of the existence of homebrew

#!/usr/bin/python
import sys, subprocess
subprocess.call(["which python"])

throws the long error

Traceback (most recent call last):
  File "installations.py", line 5, in <module>
    subprocess.call(["which python"]) 
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/subprocess.py", line 522, in call
    return Popen(*popenargs, **kwargs).wait()
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/subprocess.py", line 709, in __init__
    errread, errwrite)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/subprocess.py", line 1326, in _execute_child
    raise child_exception
OSError: [Errno 2] No such file or directory

But I know that the shell execution is working correctly on some level

#!/usr/bin/python
import sys, subprocess
subprocess.call(["whoami]) 

prints my username just fine. Am I doing something wrong, or is which for some reason not supported. Is there some better supported way of detecting the existence of an installation?

Answer 1

The failing call is trying to find an executable named 'which python' not running which with an argument of python as you likely intended. The list that you pass to call (unless shell=True is set) is the list of the command and all the arguments. Doing

subprocess.call(['which', 'python'])

will probably give you what you are looking for.

Answer 2

When calling this way you need to separate each word on your command line as a different element of an iterable:

subprocess.call(["which", "python"])

I highly recommend reading the docs several times through in order to truly understand all the possible ways of launching subprocesses and how they differ from each other.