Backoff code inside function

I am calling some functions of a lib i wanna backoff. Problem is, I cannot modify the lib.

import backoff
from requests.exceptions import TimeoutError
from somewhere import library

So I am looking for something like:

with backoff.on_exception(backoff.expo, TimeoutError, max_tries=8):
    library.some_function(yada, blah, foo, bar)
# end if

or

while backoff.on_exception(backoff.expo, TimeoutError, max_tries=8):
    library.some_function(yada, blah, foo, bar)
# end if

Not sure what is possible to do.

Edit (2017-04-27): Calling just one function can be solved like seen below

[luckydonald]

I just realized: If it is a just a single function - even in a library - you could supply the decorator yourself: So as solution to just the problem written in the original posting (calling a single function) this should be sufficient:

from somewhere.library import some_function as _some_function
import backoff
from requests.exceptions import TimeoutError

some_function = backoff.on_exception(backoff.expo, TimeoutError, max_tries=8)(_some_function)

some_function(yada, blah, foo, bar)

[luckydonald]

@pquentin, to which of my suggestion did you refer?

@bgreen-litl, What about the for+with? It also looks handy for normal code you don’t really wan’t to put in an extra function.

So 2 lines would be still less than writing it yourself every time:

for retry in backoff.retry_on_exception(backoff.expo, TimeoutError, max_tries=5):
   with retry:
     # >>> do stuff here <<<
   # end with
# end for

Now manually:

timeout = backoff.expo()
for try_count in range(5):
  try:
     # >>> do stuff here <<<
     break  # because executed successfully
  expect TimeoutError as e:
    if try_count == 5:
      raise e
    # end if
   sleep(timeout.next())
  # end try
# end for