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?
Answers:
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Method 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
Method 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.
Method 3
UPDATE 2021-12-01:
As of June 2016, the retrying package is no longer being maintained.
Consider using the active fork github.com/jd/tenacity, or alternatively github.com/litl/backoff.
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")
Method 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
Method 5
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.
Method 6
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()
Method 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
Method 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.
Method 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()
Method 10
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
Method 11
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.
Method 12
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
Method 13
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'
Method 14
You can use Python retrying package.
Retrying
It is written in Python to simplify the task of adding retry behavior to just about anything.
Method 15
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
Method 16
attempts = 3
while attempts:
try:
...
...
<status ok>
break
except:
attempts -=1
else: # executed only break was not raised
<status failed>Method 17
I use this, which can be used on any function:
def run_with_retry(func: callable, max_retries: int = 3, wait_seconds: int = 2, **func_params):
num_retries = 1
while True:
try:
return func(*func_params.values())
except Exception as e:
if num_retries > max_retries:
print('we have reached maximum errors and raising the exception')
raise e
else:
print(f'{num_retries}/{max_retries}')
print("Retrying error:", e)
num_retries += 1
sleep(wait_seconds)
Call like this:
def add(val1, val2):
return val1 + val2
run_with_retry(func=add, param1=10, param2=20)
Method 18
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
Method 19
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.
Method 20
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
Method 21
with this decorator, you can easily control errors
class catch:
def __init__(self, max=1, callback=None):
self.max = max
self.callback = callback
def set_max(self, max):
self.max = max
def handler(self, *args, **kwargs):
self.index = 0
while self.index < self.max:
self.index += 1
try:
self.func(self, *args, **kwargs)
except Exception as error:
if callable(self.callback):
self.callback(self, error, args, kwargs)
def __call__(self, func):
self.func = func
return self.handler
import time
def callback(cls, error, args, kwargs):
print('func args', args, 'func kwargs', kwargs)
print('error', repr(error), 'trying', cls.index)
if cls.index == 2:
cls.set_max(4)
else:
time.sleep(1)
@catch(max=2, callback=callback)
def test(cls, ok, **kwargs):
raise ValueError('ok')
test(1, message='hello')
Method 22
If retrying a failed attempt x number of times is what you are looking for, a single for else loop is probably what you want. Consider this example with 3 attempts:
attempts = 3
for attempt in range(1, attempts+1):
try:
if attempt < 4:
raise TypeError(f"Error raised on attempt: {attempt}")
else:
print(f'Attempt {attempt} finally worked.')
except (TypeError) as error:
print(f'Attempt {attempt} hit the exception.')
continue
else:
break
else:
print(f'Exit after final attempt: {attempt}')
print(f'nGo on to execute other code ...')
Gives the output:
Attempt 1 hit the exception. Attempt 2 hit the exception. Attempt 3 hit the exception. Exit after final attempt: 3 Go on to execute other code ...
And with one more attempt it succeeds:
attempts = 4
Gives the output:
Attempt 1 hit the exception. Attempt 2 hit the exception. Attempt 3 hit the exception. Attempt 4 finally worked. Go on to execute other code ...
Method 23
You could have a dedicated function using return to short circuit the result. For example like this:
def my_function_with_retries(..., max_retries=100):
for attempt in range(max_retries):
try:
return my_function(...)
except SomeSpecificException as error:
logging.warning(f"Retrying after failed execution: {error}")
raise SomeOtherException()
Method 24
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)
Method 25
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}')
Method 26
increment your loop variable only when the try clause succeeds
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