Lazy logger message string evaluation

Of course the following is not as efficient as a Macro:

if logger.isEnabledFor(logging.DEBUG):
    logger.debug(
        'Stupid log message ' + ' '.join([str(i) for i in range(20)])
    )

but simple, evaluates in lazy fashion and is 4 times faster than the accepted answer:

class lazyjoin:
    def __init__(self, s, items):
        self.s = s
        self.items = items

    def __str__(self):
        return self.s.join(self.items)

logger.debug(
    'Stupid log message %s', lazyjoin(' ', (str(i) for i in range(20)))
)

See benchmark-src for my setup.

The logging module already has partial support for what you want to do. Do this:

log.debug("Some message: a=%s b=%s", a, b)

... instead of this:

log.debug("Some message: a=%s b=%s" % (a, b))

The logging module is smart enough to not produce the complete log message unless the message actually gets logged somewhere.

To apply this feature to your specific request, you could create a lazyjoin class.

class lazyjoin:
    def __init__(self, s, items):
        self.s = s
        self.items = items
    def __str__(self):
        return self.s.join(self.items)

Use it like this (note the use of a generator expression, adding to the laziness):

logger.info('Stupid log message %s', lazyjoin(' ', (str(i) for i in range(20))))

Here is a demo that shows this works.

>>> import logging
>>> logging.basicConfig(level=logging.INFO)
>>> logger = logging.getLogger("log")
>>> class DoNotStr:
...     def __str__(self):
...         raise AssertionError("the code should not have called this")
... 
>>> logger.info('Message %s', DoNotStr())
Traceback (most recent call last):
...
AssertionError: the code should not have called this
>>> logger.debug('Message %s', DoNotStr())
>>>

In the demo, The logger.info() call hit the assertion error, while logger.debug() did not get that far.