python equivalent of '#define func() ' or how to comment out a function call in python

Question

my python code is interlaced with lots of function calls used for (debugging|profiling|tracing etc.) for example:

import logging

logging.root.setLevel(logging.DEBUG)
logging.debug('hello')
j = 0
for i in range(10):
    j += i
    logging.debug('i %d j %d' % (i,j))
print(j)
logging.debug('bye')

i want to #define these resource consuming functions out of the code. something like the c equivalent

#define logging.debug(val)

yes, i know the logging module logging level mechanism can be used to mask out loggings below set log level. but, im asking for a general way to have the python interpreter skip functions (that take time to run even if they dont do much)

one idea is to redefine the functions i want to comment out into empty functions:

def lazy(*args): pass
logging.debug = lazy

the above idea still calls a function, and may create a myriad of other problems

Answer 1

Python does not have a preprocessor, although you could run your python source through an external preprocessor to get the same effect - e.g. sed "/logging.debug/d" will strip out all the debug logging commands. This is not very elegant though - you will end up needing some sort of build system to run all your modules through the preprocessor and perhaps create a new directory tree of the processed .py files before running the main script.

Alternatively if you put all your debug statements in an if __debug__: block they will get optimised out when python is run with the -O (optimise) flag.

As an aside, I checked the code with the dis module to ensure that it did get optimised away. I discovered that both

if __debug__: doStuff()

and

if 0: doStuff()

are optimised, but

if False: doStuff()

is not. This is because False is a regular Python object, and you can in fact do this:

>>> False = True
>>> if False: print "Illogical, captain"
Illogical, captain

Which seems to me a flaw in the language - hopefully it is fixed in Python 3.

Edit:

This is fixed in Python 3: Assigning to True or False now gives a SyntaxError. Since True and False are constants in Python 3, it means that if False: doStuff() is now optimised:

>>> def f():
...     if False: print( "illogical")
... 
>>> dis.dis(f)
  2           0 LOAD_CONST               0 (None) 
              3 RETURN_VALUE