Using print statements only to debug

Question

The logging module has everything you could want. It may seem excessive at first, but only use the parts you need. I'd recommend using logging.basicConfig to toggle the logging level to stderr and the simple log methods, debug, info, warning, error and critical.

import logging, sys
logging.basicConfig(stream=sys.stderr, level=logging.DEBUG)
logging.debug('A debug message!')
logging.info('We processed %d records', len(processed_records))

A simple way to do this is to call a logging function:

DEBUG = True

def log(s):
    if DEBUG:
        print s

log("hello world")

Then you can change the value of DEBUG and run your code with or without logging.

The standard logging module has a more elaborate mechanism for this.

Use the logging built-in library module instead of printing.

You create a Logger object (say logger), and then after that, whenever you insert a debug print, you just put:

logger.debug("Some string")

You can use logger.setLevel at the start of the program to set the output level. If you set it to DEBUG, it will print all the debugs. Set it to INFO or higher and immediately all of the debugs will disappear.

You can also use it to log more serious things, at different levels (INFO, WARNING and ERROR).

First off, I will second the nomination of python's logging framework. Be a little careful about how you use it, however. Specifically: let the logging framework expand your variables, don't do it yourself. For instance, instead of:

logging.debug("datastructure: %r" % complex_dict_structure)

make sure you do:

logging.debug("datastructure: %r", complex_dict_structure)

because while they look similar, the first version incurs the repr() cost even if it's disabled. The second version avoid this. Similarly, if you roll your own, I'd suggest something like:

def debug_stdout(sfunc):
    print(sfunc())

debug = debug_stdout

called via:

debug(lambda: "datastructure: %r" % complex_dict_structure)

which will, again, avoid the overhead if you disable it by doing:

def debug_noop(*args, **kwargs):
    pass

debug = debug_noop

The overhead of computing those strings probably doesn't matter unless they're either 1) expensive to compute or 2) the debug statement is in the middle of, say, an n^3 loop or something. Not that I would know anything about that.