One of my logger.debug()
statements takes input that's fairly labour-intensive.
I know that I should do logger.debug("The spam is %s", spam_temperature)
rather than logger.debug("The spam is {}".format(spam_temperature))
. The problem is that the operations I need to perform to actually find out the spam_temperature
are quite CPU-intensive, and I have no use for them if the logging level is, say, INFO
.
What is the best practice in a case like this?
Found the answer myself - I added python if not logging.getLogger().isEnabledFor(logging.DEBUG): return
before the section I wanted to avoid.
(I got the inspiration from https://stackoverflow.com/a/27849836/3061818 )
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