Only printing error messages once per loop

Question

I have some Python that looks like this:

for row in reader:
    if # something
       # do things
    else:
        try:
            # do more things
        except IndexError:
            logger.info('message')

What ends up happening is I get a huge wall of output that says INFO:...message for each iteration of the loop. There’s no need for it to appear what could be over 1,000 times. I could use a ‘flag’ of sorts after I leave the loop, kind of like this:

for row in reader
    # if/else/try
        except IndexError:
            foo = True
if foo:
    logger.info('message')

But I’m wondering if there’s a more elegant way to do this. The important bits are that I do want to show the error, but only once. And I can’t break out of the loop on the error, because I need to continue processing the rest of the rows in reader. The IndexError appears when I try to create a variable from a list that might not exist. I have to do it this way because if the variable does not exist I have to skip over it instead of providing even a blank value.

Thus my weird little predicament. Is there a better way of doing this? Preferably the most ‘Pythonic’, as speed and what not isn’t such a big issue in this case.

Asked By: Eric Lagergren

||

Source

Answer 1

Just keep a counter, sometimes the simplest ways are the best. you could even add a nifty notice at the end.

bad = 0
for row in reader:
    if # something
       # do things
    else:
        try:
            # do more things
        except IndexError:
            bad += 1
            if bad == 1:
                logger.info('message')
if bad:
    logger.info('%d bad things happened' % bad)

Answered By: tdelaney

Answer 2

sys.exc_info() and sys.exc_clear() may help to store and clean the errors here. You can refer to How sys.exc_info() works? for more mechanism of these two functions.

Answered By: rzyu

Only printing error messages once per loop

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