Confusion about global variables in python

Question

I'm new to python, so please excuse what is probably a pretty dumb question.

Basically, I have a single global variable, called _debug, which is used to determine whether or not the script should output debugging information. My problem is, I can't set it in a different python script than the one that uses it.

I have two scripts:

one.py:
-------
def my_function():
 if _debug:
 print "debugging!"
two.py:
-------
from one import *
_debug = False
my_function()

Running two.py generates an error:

NameError: global name '_debug' is not defined

Can anyone tell me what I'm doing wrong?

Answer 1

There are more problems than just the leading underscore I'm afraid.

When you call my_function(), it still won't have your debug variable in its namespace, unless you import it from two.py.

Of course, doing that means you'll end up with cyclic dependencies (one.py -> two.py -> one.py), and you'll get NameErrors unless you refactor where various things are imported and declared.

One solution would be to create a simple third module which defines 'constants' like this, which can be safely imported from anywhere, e.g.:

constants.py
------------
debug = True
one.py
------
from constants import debug
#...
two.py
------
from constants import debug
#...

However, I would recommend just using the built in logging module for this - why not? It's easy to configure, simpler to use, reliable, flexible and extensible.

• unbeknown: Yes, the logging module is the way to go here.

Answer 2

Names beginning with an underscore aren't imported with

from one import *

• Bartosz Radaczyński: And it is because they mean "this is internal stuff, do not touch it, or do so at your own risk".
• James Brady: I'm actually shocked that a completely incorrect answer has been accepted and voted up so highly.
• Eli Courtwright: Well it's not completely incorrect; kgiannakakis is correct about names beginning with underscores not being imported. However, you're correct that this still isn't why his example code doesn't work. Your answer is much better.

Answer 3

You can also use the __debug__ variable for debugging. It is true if the interpreter wasn't started with the -O option. The assert statement might be helpful, too.

Answer 4

A bit more explanation: The function my_function's namespace is always in the module one. This means that when the name _debug is not found in my_function, it looks in one, not the namespace from which the function is called. Alabaster's answer provides a good solution.