Open Python IDLE and Run Command

Question

I am working with an engineering software that allows me to run python scripts internally. However, it doesn't provide me with an IDLE which I need.

I know I can store my required variables using Pickle and then load them latter. So basically I can write a script to load all variables, lets call this VarLoader.py . I also now that I can use the below to open a python idle:

import idlelib.PyShell
idlelib.PyShell.main()

But I don't know how to make it automatically run VarLoader.py in the opened IDLE. I mean something like the below would be ideal:

import idlelib.PyShell
idlelib.PyShell.main(VarLoader.py)

I also tried the below which didn't work:

import sys
sys.argv=['-c','VarLoader.py']

Any help and thoughts is appreciated.

P.S. I also know that I can do the below but I want something more professional:

import os
os.system(' '.join(['python','idle.py','-c','execfile('VarLoader.py')']))

Answer 1

With a bit of trial and error I found the solution. In case anybody in future has same problem, it can be solved as below:

import sys
sys.argv=['','-n','-t','My New Shell','-c','execfile("VarLoader.py")']
import idlelib.PyShell
idlelib.PyShell.main()