异常后如何重试？

Question

我有一个以for i in range(0, 100) 开头的循环。通常它可以正常运行，但有时会由于网络状况而失败。目前我已将其设置为在失败时，它将在 except 子句中 continue（继续到 i 的下一个数字）。

我是否可以将相同的编号重新分配给 i 并再次运行循环的失败迭代？

Answer 1

2021 年 12 月 1 日更新：

自 2016 年 6 月起，不再维护重试包。 考虑使用活动分支github.com/jd/tenacity，或者github.com/litl/backoff。

---

retrying package 是在失败时重试代码块的好方法。

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")

Answer 2

我用这个，可以用在任何功能上：

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)

这样调用：

    def add(val1, val2):
        return val1 + val2

    run_with_retry(func=add, param1=10, param2=20)

Answer 3

有了这个装饰器，你可以轻松控制错误

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')

Answer 4

我喜欢为此使用布尔值，如下所示：

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 5

装饰器是个好办法。

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

一些使用示例：

@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 6

retrying 的替代方案：tenacity 和 backoff（2020 年更新）

retrying 库以前是要走的路，但遗憾的是它有一些错误，自 2016 年以来没有任何更新。其他替代品似乎是 backoff 和 tenacity。在写这篇文章的时候，坚韧有更多的 GItHub 星（2.3k vs 1.2k）并且最近更新，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()

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

上面的代码输出如下：

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 的更多设置列于tenacity GitHub page。

Answer 7

这是我对这个问题的看法。以下retry 函数支持以下功能：

• 成功时返回被调用函数的值
• 如果尝试用尽，则引发调用函数的异常
• 尝试次数限制（0 表示无限制）
• 尝试之间的等待（线性或指数）
• 仅当异常是特定异常类型的实例时重试。
• 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)

请查看我的post 了解更多信息。

Answer 8

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

Answer 9

我最近使用我的 python 解决了这个问题，我很高兴与 stackoverflow 访问者分享它，如果需要，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}')

Answer 10

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

我的版本与上面的几个类似，但不使用单独的while 循环，如果所有重试都失败，则重新引发最新的异常。可以在顶部显式设置err = None，但不是绝对必要的，因为它应该只在出现错误时执行最终的else 块，因此设置了err。

Answer 11

我在我的代码中使用以下内容，

   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 12

如果您想要一个没有嵌套循环并在成功时调用break 的解决方案，您可以为任何可迭代开发一个快速包装retriable。这是我经常遇到的网络问题的示例 - 保存的身份验证过期。它的用法是这样的：

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 13

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()

Answer 14

您可以使用 Python 重试包。 Retrying

它是用 Python 编写的，用于简化向几乎任何事情添加重试行为的任务。

Answer 15

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

Answer 16

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

Answer 17

这里有一个和其他类似的解决方案，但是如果没有达到规定的次数或者重试，就会引发异常。

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

Answer 18

这是我关于如何解决此问题的想法：

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)

Answer 19

在您的 for 循环中执行 while True，将您的 try 代码放入其中，并仅在您的代码成功时才从该 while 循环中中断。

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

Answer 20

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

Answer 21

Python Decorator Library 中有类似的东西。

请记住，它不测试异常，而是返回值。它会重试，直到修饰函数返回 True。

稍作修改的版本应该可以解决问题。

Answer 22

我更喜欢限制重试次数，这样如果该特定项目出现问题，您最终会继续下一个项目，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.

Answer 23

不使用那些丑陋的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()

Answer 24

仅当 try 子句成功时才增加循环变量