How to find out where a Python Warning is from

You can also use the commandline to control the warnings:

python -W error::UnicodeWarning your_code.py

From the man page:

-W argument
[...] error to raise an exception instead of printing a warning message.

This will have the same effect as putting the following in your code:

import warnings
warnings.filterwarnings('error', category=UnicodeWarning)

As was already said in Andy's answer.

The most informative way to investigate a warning is to convert it into an error (Exception) so you can see its full stacktrace:

import warnings
warnings.simplefilter("error")

See warnings.

You can filter the warnings to raise which will enable you to debug (e.g. using pdb):

import warnings
warnings.filterwarnings('error')

*The warnings filter can be managed more finely (which is probably more appropriate) e.g.:

warnings.filterwarnings('error', category=UnicodeWarning)
warnings.filterwarnings('error', message='*equal comparison failed*')

Multiple filters will be looked up sequentially. ("Entries closer to the front of the list override entries later in the list, if both match a particular warning.")

If you enable logging in python, then when an exception is received you can use the method logging.exception to log when an exception has been caught - this method will print out a nicely formatted stack trace that shows you exactly in the code where the exception originated. See the python document on logging for more information.

import logging
log = logging.getLogger('my.module.logger')

try:
    return self._engine.get_loc(key)
except UnicodeWarning:
    log.exception('A log message of your choosing')

Alternatively, you can get a tuple that contains details of the exception in your code by calling sys.exc_info() (this requires you to import the sys module).