Develop Reference

Execute custom command using subprocess that is in PATH variable

I want to execute another program that I compiled earlier from python using the subprocess.call(command) function.
However, python states that it cannot find the command. I suspect that subprocess is not able to find my custom command since it does not know the PATH variable of my Ubuntu system. Is it possible to somehow execute the following code, where command is part of my PATH?
import subprocess
subprocess.run("command -args")
Running this code leads to the error command not found.

You can either provide the explicit path to your command:
subprocess.run('/full/path/to/command.sh')
or else modify your PATH variable in your Python code:
import os
os.environ['PATH'] += ':'+'/full/path/to/'
subprocess.run('command.sh')

You can modify the environment variables. But be careful when you pass arguments.
Try something like this:
import os
import subprocess
my_env = os.environ.copy()
my_env["PATH"] = "/usr/test/path:" + my_env["PATH"]
subprocess.run(["command", "-args"], env=my_env)

Python: change file/folder attributes in windows OS

I have made a folder using python3 script, and to apply multiple attributes (+h +s) to the folder I have to run ATTRIB command in Command Prompt.
But I want to know how it can be done from the same python3 script.
import os
os.makedir("C:\\AutoSC")
# Now I want the code to give the same result such that I have opned CMD and writen following command
# C:\> attrib +h +s AutoSC
# Also show in the code, necessary imported modules
I want the folder to be created and immediately hidden as system folder.
Which is not visible even after show hidden files.

Use the subprocess module or use os.system to send commands directly to OS.
import subprocess
subprocess.run(["ls","-l"])# in linux, for windows, it may change.
import os
os.system('attrib +h +s AutoSC')

Open python shell in the current directory

I want to open python shell in the current directory like cmd. My system is windows 10. How can I achieve that?
Something like that:
Suppose that I open a directory in D disk, let's say the path is "D:\PythonProjects". I tried to open cmd in current directory, in cmd I type
"python" to get python shell, but the work directory of python didn't chage.

Did you try something like this in command line?
C:> D:
D:> cd PythonProjects
D:\PythonProjects> python
[Something about Python interpreter]
>>>

You can just simply open shell
and than
import os
os.chdir(" ")
In between quotes you can specify path were you want to open that
You can also see current path by
os.getcwd()

python subprocess shouldn't run from within the folder

I am trying to use the C&C NLP library in my mac and it uses terminal as its interface. so naturally I'm trying to run the command from my python, but here's what happens:
candc:could not open model configuration file for reading:models/config
turns out candc should not be called from the same directory, and should be called from outside of the binary folder, something like "bin/candc".
how can I make this work?
this is my code:
cmd="candc/bin/candc --models models"
subprocess.check_output('{} | tee /dev/stderr'.format( cmd ), shell=True)

Pass the cwd argument with your desired working directory.
For example, if you want to run it as bin/candc from the candc directory:
import os
cmd="bin/candc --models models"
subprocess.check_output('{} | tee /dev/stderr'.format( cmd ), shell=True, cwd=os.path.abspath('candc'))
(I'm not sure whether you actually need os.path.abspath. Do test both with and without it.)

Use the full path in cmd:
cmd = "/home/your-username/python-programs/cnc/candc/bin/canc --models models
Whatever that full path might be. You can use (if you're on linux) pwd inside the candc directory to find out what it is.

open(file) from anywhere

Working on OS X Lion, I'm trying to open a file in my python-program from anywhere in the terminal. I have set the following function in my .bash_profile:
function testprogram() {python ~/.folder/.testprogram.py}
This way I can(in the terminal) run my testprogram from a different directory than my ~/.
Now, if I'm in my home directory, and run the program, the following would work
infile = open("folder2/test.txt", "r+")
However, if I'm in a different directory from my home-folder and write "testprogram" in the terminal, the program starts but is unable to find the file test.txt.
Is there any way to always have python open the file from the same location unaffected of where i run the program from?

If you want to make it multiplatform I would recommend
import os
open(os.path.join(os.path.expanduser('~'),'rest/of/path/to.file'))

Use the tilde to represent the home folder, just as you would in the .bash_profile, and use os.path.expanduser.
import os
infile = open(os.path.expanduser("~/folder2/test.txt"), "r+")

This is no different than referring to a file via anything else besides Python, you need to use an absolute path rather than a relative one.
If you'd like to refer to files relative to the location of the script, you can also use the __file__ attribute in your module to get the location of the currently running module.
Also, rather than using a shell function, just give your script a shebang line (#!/usr/bin/env python), chmod +x it, and put it someplace on your PATH.

You could simply run the function in a subshell so you can cd home before running it:
function testprogram() { (
cd && python .folder/.testprogram.py
) }

In python sys.argv[0] will be set to the path of the script (if known). You can use this plus the functions in os.path to access files relative to the directory containing the script. For example
import sys, os.path
script_path = sys.argv[0]
script_dir = os.path.dirname(script_path)
def script_relative(filename):
return os.path.join(script_dir, filename)
infile = open(script_relative("folder2/test.txt"), "r+")
David Robinson points out that sys.argv[0] may not be a full pathname. Instead of using sys.argv[0] you can use __file__ if this is the case.

Use either full path like
/path/to/my/folder2/test.txt
or abbreviated one
~/folder2/test.txt