Several ways.

• From the shell

  python someFile.py
  

• From inside IDLE, hit F5.

• If you're typing interactively, try this (Python3):

  >>> exec(open("filename.py").read())
  

• For Python 2:

  >>> variables= {}
  >>> execfile( "someFile.py", variables )
  >>> print variables # globals from the someFile module
  

execfile executes the script in the local namespace. You could simply assign the value to sys.argv in a prior call to exec:

PythonInterpreter interpreter = new PythonInterpreter();
interpreter.exec(
    "import sys\n"
    +"sys.argv = ['Foo', 'Bar']");
interpreter.execfile("J:/test.py");

Where the script is:

import sys

print(sys.argv)

prints:

['Foo', 'Bar']

I looked into the question of your comment, and it looks like you would need to set python.path in a Properties object that you then pass to PythonInterpreter.initialize. You could also use this to pass the arguments:

Properties p = new Properties();
p.setProperty("python.path", "J:/WS/jython"); // Sets the module path

PythonInterpreter.initialize(System.getProperties(), p, new String[]{ "Foo", "Bar" });

PythonInterpreter interpreter = new PythonInterpreter();
interpreter.execfile("J:/WS/jython/main.py");

According to the Python documentation, instead of this:

execfile("./filename") 

Use this:

exec(open("./filename").read())

See Python's docs for:

• What’s New In Python 3.0
• execfile
• exec

Actually, wouldn't we want to do this?

import sys
sys.argv = ['abc.py','arg1', 'arg2']
execfile('abc.py')