[Python-ideas] Retrying EAFP without DRY

Tue Jan 24 15:06:23 CET 2012

On Tue, Jan 24, 2012 at 9:36 PM, Paul Moore <p.f.moore at gmail.com> wrote:
> A construct that let end users abstract this type of pattern would
> probably be a far bigger win than a retry statement. (And it may be
> that it has the benefit of already existing, I just couldn't see it
> :-))

You just need to move the pause inside the iterator:
 def backoff(attempts, first_delay, scale=2):
 delay = first_delay
 for attempt in range(1, attempts+1):
 yield attempt
 time.sleep(delay)
 delay *= 2
 for __ in backoff(MAX_ATTEMPTS, 5):
 try:
 response = urllib2.urlopen(url)
 except urllib2.HTTPError as e:
 if e.code == 503: # Service Unavailable.
 continue
 raise
 break
You can also design smarter versions where the object yielded is
mutable, making it easy to pass state back into the iterator.
Cheers,
Nick.
-- 
Nick Coghlan   |   ncoghlan at gmail.com   |   Brisbane, Australia