python - How to retry after exception?

Question

I have a loop starting with for i in range(0, 100). Normally it runs correctly, but sometimes it fails due to network conditions. Currently I have it set so that on failure, it will continue in the except clause (continue on to the next number for i).

Is it possible for me to reassign the same number to i and run through the failed iteration of the loop again?

Answer 1

Do a while True inside your for loop, put your try code inside, and break from that while loop only when your code succeeds.

for i in range(0,100):
    while True:
        try:
            # do stuff
        except SomeSpecificException:
            continue
        break

Answer 2

I prefer to limit the number of retries, so that if there's a problem with that specific item you will eventually continue onto the next one, thus:

for i in range(100):
  for attempt in range(10):
    try:
      # do thing
    except:
      # perhaps reconnect, etc.
    else:
      break
  else:
    # we failed all the attempts - deal with the consequences.

Answer 3

The retrying package is a nice way to retry a block of code on failure.

For example:

@retry(wait_random_min=1000, wait_random_max=2000)
def wait_random_1_to_2_s():
    print("Randomly wait 1 to 2 seconds between retries")

Answer 4

Here is a solution similar to others, but it will raise the exception if it doesn't succeed in the prescribed number or retries.

tries = 3
for i in range(tries):
    try:
        do_the_thing()
    except KeyError as e:
        if i < tries - 1: # i is zero indexed
            continue
        else:
            raise
    break

Answer 5

The more "functional" approach without using those ugly while loops:

def tryAgain(retries=0):
    if retries > 10: return
    try:
        # Do stuff
    except:
        retries+=1
        tryAgain(retries)

tryAgain()

Answer 6

Alternatives to retrying: tenacity and backoff (2020 update)

The retrying library was previously the way to go, but sadly it has some bugs and it hasn't got any updates since 2016. Other alternatives seem to be backoff and tenacity. During the time of writing this, the tenacity had more GItHub stars (2.3k vs 1.2k) and was updated more recently, hence I chose to use it. Here is an example:

from functools import partial
import random # producing random errors for this example

from tenacity import retry, stop_after_delay, wait_fixed, retry_if_exception_type

# Custom error type for this example
class CommunicationError(Exception):
    pass

# Define shorthand decorator for the used settings.
retry_on_communication_error = partial(
    retry,
    stop=stop_after_delay(10),  # max. 10 seconds wait.
    wait=wait_fixed(0.4),  # wait 400ms 
    retry=retry_if_exception_type(CommunicationError),
)()


@retry_on_communication_error
def do_something_unreliable(i):
    if random.randint(1, 5) == 3:
        print('Run#', i, 'Error occured. Retrying.')
        raise CommunicationError()

for i in range(100):
    do_something_unreliable(i)

The above code outputs something like:

Run# 3 Error occured. Retrying.
Run# 5 Error occured. Retrying.
Run# 6 Error occured. Retrying.
Run# 6 Error occured. Retrying.
Run# 10 Error occured. Retrying.
.
.
.

More settings for the tenacity.retry are listed on the tenacity GitHub page.

Answer 7

The clearest way would be to explicitly set i. For example:

i = 0
while i < 100:
    i += 1
    try:
        # do stuff

    except MyException:
        continue

Answer 8

for _ in range(5):
    try:
        # replace this with something that may fail
        raise ValueError("foo")

    # replace Exception with a more specific exception
    except Exception as e:
        err = e
        continue

    # no exception, continue remainder of code
    else:
        break

# did not break the for loop, therefore all attempts
# raised an exception
else:
    raise err

My version is similar to several of the above, but doesn't use a separate while loop, and re-raises the latest exception if all retries fail. Could explicitly set err = None at the top, but not strictly necessary as it should only execute the final else block if there was an error and therefore err is set.

Answer 9

A generic solution with a timeout:

import time

def onerror_retry(exception, callback, timeout=2, timedelta=.1):
    end_time = time.time() + timeout
    while True:
        try:
            yield callback()
            break
        except exception:
            if time.time() > end_time:
                raise
            elif timedelta > 0:
                time.sleep(timedelta)

Usage:

for retry in onerror_retry(SomeSpecificException, do_stuff):
    retry()

Answer 10

There is something similar in the Python Decorator Library.

Please bear in mind that it does not test for exceptions, but the return value. It retries until the decorated function returns True.

A slightly modified version should do the trick.

Answer 11

Using while and a counter:

count = 1
while count <= 3:  # try 3 times
    try:
        # do_the_logic()
        break
    except SomeSpecificException as e:
        # If trying 3rd time and still error?? 
        # Just throw the error- we don't have anything to hide :)
        if count == 3:
            raise
        count += 1

Answer 12

Using recursion

for i in range(100):
    def do():
        try:
            ## Network related scripts
        except SpecificException as ex:
            do()
    do() ## invoke do() whenever required inside this loop

Answer 13

You can use Python retrying package. Retrying

It is written in Python to simplify the task of adding retry behavior to just about anything.

Answer 14

I use following in my codes,

   for i in range(0, 10):
    try:
        #things I need to do
    except ValueError:
        print("Try #{} failed with ValueError: Sleeping for 2 secs before next try:".format(i))
        time.sleep(2)
        continue
    break

Answer 15

attempts = 3
while attempts:
  try:
     ...
     ...
     <status ok>
     break
  except:
    attempts -=1
else: # executed only break was not  raised
   <status failed>

Answer 16

Decorator is a good approach.

from functools import wraps
import time

class retry:
    def __init__(self, success=lambda r:True, times=3, delay=1, raiseexception=True, echo=True):
        self.success = success
        self.times = times
        self.raiseexception = raiseexception
        self.echo = echo
        self.delay = delay
    def retry(fun, *args, success=lambda r:True, times=3, delay=1, raiseexception=True, echo=True, **kwargs):
        ex = Exception(f"{fun} failed.")
        r = None
        for i in range(times):
            if i > 0:
                time.sleep(delay*2**(i-1))
            try:
                r = fun(*args, **kwargs)
                s = success(r)
            except Exception as e:
                s = False
                ex = e
                # raise e
            if not s:
                continue
            return r
        else:
            if echo:
                print(f"{fun} failed.", "args:", args, kwargs, "\nresult: %s"%r)
            if raiseexception:
                raise ex
    def __call__(self, fun):
        @wraps(fun)
        def wraper(*args, retry=0, **kwargs):
            retry = retry if retry>0 else self.times
            return self.__class__.retry(fun, *args, 
                                        success=self.success, 
                                        times=retry,
                                        delay=self.delay,
                                        raiseexception = self.raiseexception,
                                        echo = self.echo,
                                        **kwargs)
        return wraper

some usage examples:

@retry(success=lambda x:x>3, times=4, delay=0.1)
def rf1(x=[]):
    x.append(1)
    print(x)
    return len(x)
> rf1()

[1]
[1, 1]
[1, 1, 1]
[1, 1, 1, 1]

4

@retry(success=lambda x:x>3, times=4, delay=0.1)
def rf2(l=[], v=1):
    l.append(v)
    print(l)
    assert len(l)>4
    return len(l)
> rf2(v=2, retry=10) #overwite times=4

[2]
[2, 2]
[2, 2, 2]
[2, 2, 2, 2]
[2, 2, 2, 2, 2]

5

> retry.retry(lambda a,b:a+b, 1, 2, times=2)

3

> retry.retry(lambda a,b:a+b, 1, "2", times=2)

TypeError: unsupported operand type(s) for +: 'int' and 'str'

Answer 17

If you want a solution without nested loops and invoking break on success you could developer a quick wrap retriable for any iterable. Here's an example of a networking issue that I run into often - saved authentication expires. The use of it would read like this:

client = get_client()
smart_loop = retriable(list_of_values):

for value in smart_loop:
    try:
        client.do_something_with(value)
    except ClientAuthExpired:
        client = get_client()
        smart_loop.retry()
        continue
    except NetworkTimeout:
        smart_loop.retry()
        continue

Answer 18

Here is my take on this issue. The following retry function supports the following features:

• Returns the value of the invoked function when it succeeds
• Raises the exception of the invoked function if attempts exhausted
• Limit for the number of attempts (0 for unlimited)
• Wait (linear or exponential) between attempts
• Retry only if the exception is an instance of a specific exception type.
• Optional logging of attempts

import time

def retry(func, ex_type=Exception, limit=0, wait_ms=100, wait_increase_ratio=2, logger=None):
    attempt = 1
    while True:
        try:
            return func()
        except Exception as ex:
            if not isinstance(ex, ex_type):
                raise ex
            if 0 < limit <= attempt:
                if logger:
                    logger.warning("no more attempts")
                raise ex

            if logger:
                logger.error("failed execution attempt #%d", attempt, exc_info=ex)

            attempt += 1
            if logger:
                logger.info("waiting %d ms before attempt #%d", wait_ms, attempt)
            time.sleep(wait_ms / 1000)
            wait_ms *= wait_increase_ratio

Usage:

def fail_randomly():
    y = random.randint(0, 10)
    if y < 10:
        y = 0
    return x / y


logger = logging.getLogger()
logger.setLevel(logging.INFO)
logger.addHandler(logging.StreamHandler(stream=sys.stdout))

logger.info("starting")
result = retry.retry(fail_randomly, ex_type=ZeroDivisionError, limit=20, logger=logger)
logger.info("result is: %s", result)

See my post for more info.

Answer 19

I like to use bool values for this, like so:

success = False
num_try = 0
while success is False:
    if num_try >= 10: # or any number
        # handle error how  you please
    try:
        # code
        success = True
    except Exception as e:
        # record or do something with exception if needed
        num_try += 1

Answer 20

i recently worked with my python on a solution to this problem and i am happy to share it with stackoverflow visitors please give feedback if it is needed.

print("\nmonthly salary per day and year converter".title())
print('==' * 25)


def income_counter(day, salary, month):
    global result2, result, is_ready, result3
    result = salary / month
    result2 = result * day
    result3 = salary * 12
    is_ready = True
    return result, result2, result3, is_ready


i = 0
for i in range(5):
    try:
        month = int(input("\ntotal days of the current month: "))
        salary = int(input("total salary per month: "))
        day = int(input("Total Days to calculate> "))
        income_counter(day=day, salary=salary, month=month)
        if is_ready:
            print(f'Your Salary per one day is: {round(result)}')
            print(f'your income in {day} days will be: {round(result2)}')
            print(f'your total income in one year will be: {round(result3)}')
            break
        else:
            continue
    except ZeroDivisionError:
        is_ready = False
        i += 1
        print("a month does'nt have 0 days, please try again")
        print(f'total chances left: {5 - i}')
    except ValueError:
        is_ready = False
        i += 1
        print("Invalid value, please type a number")
        print(f'total chances left: {5 - i}')

Answer 21

Here's my idea on how to fix this:

j = 19
def calc(y):
    global j
    try:
        j = j + 8 - y
        x = int(y/j)   # this will eventually raise DIV/0 when j=0
        print("i = ", str(y), " j = ", str(j), " x = ", str(x))
    except:
        j = j + 1   # when the exception happens, increment "j" and retry
        calc(y)
for i in range(50):
    calc(i)

Answer 22

increment your loop variable only when the try clause succeeds