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Groups > comp.lang.python > #24621 > unrolled thread
| Started by | andrea crotti <andrea.crotti.0@gmail.com> |
|---|---|
| First post | 2012-06-28 17:26 +0100 |
| Last post | 2012-06-29 05:45 -0700 |
| Articles | 6 — 4 participants |
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Re: retry many times decorator andrea crotti <andrea.crotti.0@gmail.com> - 2012-06-28 17:26 +0100
Re: retry many times decorator Steven D'Aprano <steve+comp.lang.python@pearwood.info> - 2012-06-28 17:43 +0000
Re: retry many times decorator Andrea Crotti <andrea.crotti.0@gmail.com> - 2012-06-28 22:11 +0100
Re: retry many times decorator andrea crotti <andrea.crotti.0@gmail.com> - 2012-06-29 09:53 +0100
Re: retry many times decorator Justin Barber <barber.justin@gmail.com> - 2012-06-29 05:45 -0700
Re: retry many times decorator Justin Barber <barber.justin@gmail.com> - 2012-06-29 05:45 -0700
| From | andrea crotti <andrea.crotti.0@gmail.com> |
|---|---|
| Date | 2012-06-28 17:26 +0100 |
| Subject | Re: retry many times decorator |
| Message-ID | <mailman.1612.1340900799.4697.python-list@python.org> |
> Returning a boolean isn't very Pythonic. It would be better, IMHO, if
> it could swallow a specified exception (or specified exceptions?)
> raised when an attempt failed, up to the maximum permitted number of
> attempts. If the final attempt fails, propagate the exception.
> --
> http://mail.python.org/mailman/listinfo/python-list
Yes right I also didn't like it.. Now it become something as below,
so I capture every possible exception (since it must be generic) and
log what exception was raised. I'm not re-raising because if it fails
and it's fatal I should just exit, and if it's not fatal it should
just continue..
class retry_n_times:
def __init__(self, ntimes=3, timeout=3, fatal=True):
self.ntimes = ntimes
self.timeout = timeout
self.fatal = fatal
def __call__(self, func):
def _retry_n_times(*args, **kwargs):
attempts = 0
while True:
logger.debug("Attempt number %s of %s" % (attempts,
func.__name__))
try:
ret = func(*args, **kwargs)
except Exception as e:
logger.error("Got exception %s with error %s" %
(type(e), str(e)))
sleep(self.timeout)
else:
return ret
attempts += 1
if attempts == self.ntimes:
logger.error("Giving up the attempts while running
%s" % func.__name__)
if self.fatal:
exit(100)
return _retry_n_times
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| From | Steven D'Aprano <steve+comp.lang.python@pearwood.info> |
|---|---|
| Date | 2012-06-28 17:43 +0000 |
| Message-ID | <4fec97d9$0$29978$c3e8da3$5496439d@news.astraweb.com> |
| In reply to | #24621 |
On Thu, 28 Jun 2012 17:26:36 +0100, andrea crotti wrote: >> Returning a boolean isn't very Pythonic. It would be better, IMHO, if >> it could swallow a specified exception (or specified exceptions?) >> raised when an attempt failed, up to the maximum permitted number of >> attempts. If the final attempt fails, propagate the exception. -- >> http://mail.python.org/mailman/listinfo/python-list > > > Yes right I also didn't like it.. Now it become something as below, so > I capture every possible exception (since it must be generic) I disagree. If you make a coding error in your function, why do you think it is useful to retry that buggy code over and over again? It's never going to get less buggy unless you see the exception and fix the bug. For any operation that you want to retry, identify the *temporary* errors, catch them, and retry the request. *Permanent* errors should immediately fail, without retrying. *Unexpected* errors should not be caught, since they probably represent a bug in your code. E.g. when trying to read a HTTP resource, there is no point retrying a 400 Bad Request repeatedly, and it is rather anti-social to keep hammering a website when it's already told you that your syntax is wrong. Likewise, if you get 401, there is no point in retrying with the exact same request -- you need a *different* request, this time using authentication. A 403 is a permanent error and there's no point in retrying. A 404 might be temporary (although it probably isn't). A 503 or 504 is likely to be temporary. And so forth. http://en.wikipedia.org/wiki/List_of_HTTP_status_codes And if you get an unexpected AttributeError: 'NoneType' object has no attribute 'read' (as I just did today, in one of my scripts), that's a bug that should be fixed. Almost any other resource you might try is likely to have similar responses: some errors will indicate temporary failures, others will be permanent, and some will indicate a bug in your code. I think it is rude and pointless to repeatedly retry the same failed request if it cannot possibly succeed. I also find that exponential backoff for the delay is best. E.g. wait 1 second after the first failed attempt, 2 seconds after the second, 4 seconds after the third, 8 seconds after the fourth, etc. -- Steven
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| From | Andrea Crotti <andrea.crotti.0@gmail.com> |
|---|---|
| Date | 2012-06-28 22:11 +0100 |
| Message-ID | <mailman.1621.1340917948.4697.python-list@python.org> |
| In reply to | #24628 |
On 06/28/2012 06:43 PM, Steven D'Aprano wrote: > On Thu, 28 Jun 2012 17:26:36 +0100, andrea crotti wrote: > > > I disagree. If you make a coding error in your function, why do you think > it is useful to retry that buggy code over and over again? It's never > going to get less buggy unless you see the exception and fix the bug. > > For any operation that you want to retry, identify the *temporary* > errors, catch them, and retry the request. *Permanent* errors should > immediately fail, without retrying. *Unexpected* errors should not be > caught, since they probably represent a bug in your code. Ah well maybe I wasn't clear, but I'm not going to retry random things, I will only decorate the functions that I know for sure that could go wrong for temporary network problems. For example they told me that sometimes mysql just doesn't respond in time for some reasons, but there's nothing permanently wrong, so retrying is the best option.. It would be good of course, however, to catch the exceptions that are known to be permanent problems in the function at least, and leave the retry as last resource.. Thanks for the idea of the exponential backoff, which is also a better name than timeout for the variable..
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| From | andrea crotti <andrea.crotti.0@gmail.com> |
|---|---|
| Date | 2012-06-29 09:53 +0100 |
| Message-ID | <mailman.1628.1340960024.4697.python-list@python.org> |
| In reply to | #24628 |
On the other hand now that I think again even supposing there is a permanent error like MySql completely down, retrying continuosly won't do any harm anyway because the machine will not be able to do anything else anyway, when someone will fix MySql it would restart again without human intervention. So I think I could even just let it retry and use maybe a SMTPHanlder for the logging errors, to make the notification of problems very quick..
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| From | Justin Barber <barber.justin@gmail.com> |
|---|---|
| Date | 2012-06-29 05:45 -0700 |
| Message-ID | <mailman.1629.1340973905.4697.python-list@python.org> |
| In reply to | #24650 |
On Friday, June 29, 2012 4:53:43 AM UTC-4, andrea crotti wrote: > On the other hand now that I think again even supposing there is a > permanent error like MySql completely down, retrying continuosly > won't do any harm anyway because the machine will not be able to do > anything else anyway, when someone will fix MySql it would > restart again without human intervention. > > So I think I could even just let it retry and use maybe a SMTPHanlder > for the logging errors, to make the notification of problems very > quick.. Rather then write a decorator, sounds like you should write your own class or method connecting to the database that has the ability to retry. Are you finding you are decorating only db calls or are you decorating all sorts of stuff?
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| From | Justin Barber <barber.justin@gmail.com> |
|---|---|
| Date | 2012-06-29 05:45 -0700 |
| Message-ID | <01e8fd76-3cdb-4ffc-9553-cd2d95d3c260@googlegroups.com> |
| In reply to | #24650 |
On Friday, June 29, 2012 4:53:43 AM UTC-4, andrea crotti wrote: > On the other hand now that I think again even supposing there is a > permanent error like MySql completely down, retrying continuosly > won't do any harm anyway because the machine will not be able to do > anything else anyway, when someone will fix MySql it would > restart again without human intervention. > > So I think I could even just let it retry and use maybe a SMTPHanlder > for the logging errors, to make the notification of problems very > quick.. Rather then write a decorator, sounds like you should write your own class or method connecting to the database that has the ability to retry. Are you finding you are decorating only db calls or are you decorating all sorts of stuff?
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