我有一个循环,开头为for I,范围为(0,100)。正常情况下,它可以正常运行,但有时由于网络条件而出现故障。目前,我已经设置它,以便在失败时,它将继续在except子句中(继续到I的下一个数字)。

我是否可以将相同的数字重新分配给i,并再次运行失败的循环迭代?


当前回答

Decorator是一个很好的方法。

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

一些用法示例:

@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'

其他回答

这里有一个快速装饰器来处理这个问题。7行,没有依赖关系。

def retry(exception=Exception, retries=3, delay=0):
    def wrap(func):
        for i in range(retries):
            try:
                return func()
            except exception as e:
                print(f'Retrying {func.__name__}: {i}/{retries}')
                time.sleep(delay)
        raise e
    return wrap

@retry()
def do_something():
  ...
@retry(HTTPError, retries=100, delay=3)
def download_something():
  ...

可以添加的一个功能是扩展异常以处理多个异常(splat一个列表)。

Decorator是一个很好的方法。

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

一些用法示例:

@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'

在for循环中执行while True,将try代码放入其中,只有当代码成功时才退出while循环。

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

我倾向于限制重试次数,这样如果某个特定项目出现问题,你就可以继续进行下一个项目,如下:

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.
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,但没有使用单独的while循环,如果所有重试都失败,则重新引发最新的异常。可以显式地在顶部设置err = None,但不是严格必要的,因为它只应该在出现错误时执行最后一个else块,因此设置了err。