Python: how to do lazy debug logging

Question

I have some python like this:

def foo():
    logger = logging.getLogger()
    # do something here
    logger.debug(\'blah blah {}\'.format(expensive_func()))         


        

Answer 1

As Vinay Sajip suggests, you can do the following:

def foo():
    logger = logging.getLogger()
    if logger.isEnabledFor(logging.DEBUG):
        logger.debug('blah blah {}'.format(expensive_func()))
        logger.debug('Message: {}'.format(expf_1(expf_2(some_arg))))
        logger.debug('Message: {}', Lazy(expf_1, Lazy(expf_2, some_arg)))
foo()

Which is already lazy!

That's because the then-expressions

        logger.debug('blah blah {}'.format(expensive_func()))
        logger.debug('Message: {}'.format(expf_1(expf_2(some_arg))))
        logger.debug('Message: {}', Lazy(expf_1, Lazy(expf_2, some_arg)))

are only evaluated if and only if logger.isEnabledFor(logging.DEBUG) returns True, i.e. if and only if their evaluation is needed.

---

Even more

logging.info(DeferredMessage(expensive_func, 1, 2))

is not as lazy as one may think: DeferredMessage(expensive_func, 1, 2) have to be evaluated in an eager fashion. Which is in addition slower than evaluating:

    if logger.isEnabledFor(logging.DEBUG):