[Solved] Easier way to enable verbose logging

I want to add a debug print statement test, if I enable --verbose from the command line and if I have the following in the script.


I went through the following questions, but couldn’t get the answer…

Solution #1:

You need to combine the wisdom of the Argparse Tutorial with Python’s Logging HOWTO. Here’s an example…

> cat verbose.py 
#!/usr/bin/env python

import argparse
import logging

parser = argparse.ArgumentParser(
    description='A test script for http://stackoverflow.com/q/14097061/78845'
parser.add_argument("-v", "--verbose", help="increase output verbosity",

args = parser.parse_args()
if args.verbose:

logging.debug('Only shown in debug mode')

Run the help:

> ./verbose.py -h
usage: verbose.py [-h] [-v]

A test script for http://stackoverflow.com/q/14097061/78845

optional arguments:
  -h, --help     show this help message and exit
  -v, --verbose  increase output verbosity

Running in verbose mode:

> ./verbose.py -v
DEBUG:root:Only shown in debug mode

Running silently:

> ./verbose.py   
Respondent: Johnsyweb

Solution #2:

I find both --verbose (for users) and --debug (for developers) useful. Here’s how I do it with logging and argparse:

import argparse
import logging

parser = argparse.ArgumentParser()
    '-d', '--debug',
    help="Print lots of debugging statements",
    action="store_const", dest="loglevel", const=logging.DEBUG,
    '-v', '--verbose',
    help="Be verbose",
    action="store_const", dest="loglevel", const=logging.INFO,
args = parser.parse_args()    

So if --debug is set, the logging level is set to DEBUG. If --verbose, logging is set to INFO. If neither, the lack of --debug sets the logging level to the default of WARNING.

Respondent: Matthew Leingang

Solution #3:

Here is a more concise method, that does bounds checking, and will list valid values in help:

parser = argparse.ArgumentParser(description='This is a demo.')
parser.add_argument("-l", "--log", dest="logLevel", choices=['DEBUG', 'INFO', 'WARNING', 'ERROR', 'CRITICAL'], help="Set the logging level")

args = parser.parse_args()
if args.logLevel:
    logging.basicConfig(level=getattr(logging, args.logLevel))


demo.py --log DEBUG
Respondent: Stickley

Solution #4:

Another variant would be to count the number of -v and use the count as an index to the a list with the actual levels from logging:

import argparse
import logging

parser = argparse.ArgumentParser()
parser.add_argument('-v', '--verbose', action='count', default=0)
args = parser.parse_args()

levels = [logging.WARNING, logging.INFO, logging.DEBUG]
level = levels[min(len(levels)-1,args.verbose)]  # capped to number of levels

                    format="%(asctime)s %(levelname)s %(message)s")

logging.debug("a debug message")
logging.info("a info message")
logging.warning("a warning message")

This works for -vvvv, -vvv, -vv, -v, -v -v , etc, If no -v then logging.WARNING is selected if more -v are provided it will step to INFO and DEBUG

Respondent: RubenLaguna

Solution #5:

You can explicity specify a level as an integer after the -v flag:

parser = argparse.ArgumentParser()
parser.add_argument("-v", "--verbose", const=1, default=0, type=int, nargs="?",
                    help="increase verbosity: 0 = only warnings, 1 = info, 2 = debug. No number means info. Default is no verbosity.")
args = parser.parse_args()

logger = logging.getLogger()
if args.verbose == 0:
elif args.verbose == 1:
elif args.verbose == 2:
Respondent: Stephan

Solution #6:

Here’s another take on having argparse count the -v option to increase verbosity up two levels from the default WARNING to INFO (-v) to DEBUG (-vv). This does not map to the constants defined by logging but rather calculates the value directly, limiting the input:

print( "Verbosity / loglevel:", args.v )
logging.basicConfig( level=10*(3-max(0,min(args.v,3))) )
logging.debug("debug") # 10
logging.info("info") # 20
logging.warning("warning") # 30 - The default level is WARNING, which means that only events of this level and above will be tracked
logging.error("error") # 40
logging.critical("critical") # 50
Respondent: handle

Solution #7:

if you want to enable logging.DEBUG level for a script you don’t want to (or cannot) edit, you can customize your startup:

[email protected]:~$ python -c "import site; site._script()"
USER_BASE: '/home/jcomeau/.local' (exists)
USER_SITE: '/home/jcomeau/.local/lib/python2.7/site-packages' (exists)
[email protected]:~$ mkdir -p ~/.local/lib/python2.7/site-packages
[email protected]:~$ vi ~/.local/lib/python2.7/site-packages/usercustomize.py

enter the following:

import os, logging
if os.getenv('DEBUGGING'):
    logging.basicConfig(level = logging.DEBUG)

then you can just:

[email protected]:~$ mkdir -p /tmp/some/random/
[email protected]:~$ echo 'import logging; logging.debug("test")' >> /tmp/some/random/script.py
[email protected]:~$ DEBUGGING=1 python /tmp/some/random/script.py 

from Paul Ollis at http://nedbatchelder.com/blog/201001/running_code_at_python_startup.html

2017-07-18: I’ve since switched to a different method:

logging.basicConfig(level=logging.DEBUG if __debug__ else logging.INFO)

what this does is, if you’re running without optimization (as in python script.py) you get the DEBUG-level stuff, whereas if you run with python -OO script.py you don’t. no environment variables to set.

Respondent: jcomeau_ictx

The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 .

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