conditional evaluation of source file in python

Say I have a file that's only used in pre-production code

I want to ensure it gets not run in production code- any calls out to it have to fail.

This snippet at the top of the file doesn't work - it breaks the Python grammar, which specifies that return must take place in a function.

if not __debug__:
   return None

What is the best solution here - the one that doesn't involve开发者_如何学运维 making a gigantic else, that is. :-)

if not __debug__:
    raise RuntimeError('This module must not be run in production code.')

Maybe split the non-production code out into a module which is conditionally imported from the main code?

if __debug__:
    import non_production
    non_production.main()

Updated: Based on your comment, you might want to look a 3rd-party library pypreprocessor which lets you do C-style preprocessor directives in Python. They provide a debugging example which seems very close to what you're looking for (ignoring inline debug code without requiring indentation).

Copy/pasted from that url:

from pypreprocessor import pypreprocessor
pypreprocessor.parse()
#define debug

#ifdef debug
print('The source is in debug mode')
#else
print('The source is not in debug mode')
#endif

import sys

if not __debug__:
    sys.exit()

Documentation for sys.exit.

One way you can do this is to hide all of the stuff in that module in another module that is imported conditionally.

.
├── main.py
├── _test.py
├── test.py

main.py:

import test
print dir(test)

test.py:

if __debug__:
    from _test import *

_test.py:

a = 1
b = 2

Edit:

Just realized your comment in another answer where you said "I am hoping to avoid creating two different files for what amounts to an #ifdef". As shown in another answer, there really isn't any way to do what you want without an if statement.

I've upvoted the answer by samplebias, as I think that answer (plus edit) describes the closest you're going to get without using an if statement.