异常后如何重试?

问题:异常后如何重试?

我有一个以开头的循环for i in range(0, 100)。正常情况下,它可以正常运行,但有时由于网络条件而失败。目前,我已对其进行了设置,以便在失败时,它将continue在except子句中(继续到的下一个数字i)。

我是否可以将相同的数字重新分配给i循环并再次执行失败的循环?

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?


回答 0

做一个while True内部的for循环,把你的try代码中,并突破从while只有当你的代码的成功循环。

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

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

回答 1

我更喜欢限制重试的次数,这样,如果该特定项目有问题,您最终将继续进行下一个,因此:

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.

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.

回答 2

重试包是重试的代码块失败的好办法。

例如:

@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")

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")

回答 3

这是一个与其他解决方案类似的解决方案,但是如果未按规定的次数或重试次数失败,则会引发异常。

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

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

回答 4

没有使用那些难看的while循环的更“实用”的方法:

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

tryAgain()

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()

回答 5

最清晰的方法是显式设置i。例如:

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

    except MyException:
        continue

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

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

    except MyException:
        continue

回答 6

带有超时的通用解决方案:

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)

用法:

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

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()

回答 7

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

我的版本与上述几种类似,但是没有使用单独的while循环,并且如果所有重试均失败,则会重新引发最新的异常。可以err = None在顶部显式设置,但不是严格必需的,因为只有在else出现错误并因此err被设置时才执行最后一个块。

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.


回答 8

Python Decorator库中有类似的东西

请记住,它不会测试异常,而是返回值。重试,直到修饰的函数返回True。

稍加修改的版本应该可以解决问题。

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.


回答 9

使用递归

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

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

回答 10

使用while和计数器:

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

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

回答 11

您可以使用Python重试包。 重试

它是用Python编写的,以简化将重试行为添加到几乎所有内容的任务。

You can use Python retrying package. Retrying

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


回答 12

retryingtenacitybackoff(2020更新)的替代方案

重试库是以前的路要走,但可悲的是它有一些缺陷,自2016年其他选择似乎还没有得到任何更新补偿坚韧。在撰写本文时,坚韧程度更高的GItHub星星(2.3k和1.2k)并已更新,因此我选择使用它。这是一个例子:

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()

上面的代码输出类似:

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.
.
.
.

tenacity.retry韧度GitHub页面上列出了的更多设置。

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.


回答 13

如果您想要一个没有嵌套循环且无需break成功调用的解决方案,则可retriable以为任何可迭代的对象开发一个快速包装。这是我经常遇到的网络问题的示例-保存的身份验证过期。它的使用将如下所示:

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

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

回答 14

我在代码中使用以下代码,

   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

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

回答 15

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

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

回答 16

这是我对这个问题的看法。以下retry功能支持以下功能:

  • 成功时返回被调用函数的值
  • 如果尝试已用尽,则引发被调用函数的异常
  • 尝试次数的限制(0表示无限制)
  • 尝试之间等待(线性或指数)
  • 仅当异常是特定异常类型的实例时才重试。
  • 可选的尝试记录
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

用法:

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)

有关更多信息,请参见我的帖子

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.


回答 17

这是我的解决方法:

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)

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)

回答 18

我最近与我的python合作解决了这个问题,很高兴与stackoverflow访问者分享它,如果需要的话请提供反馈。

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}')

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}')

回答 19

仅在try子句成功时增加循环变量

increment your loop variable only when the try clause succeeds