Python 是否可以在不考虑异常的情况下重试特定代码段？

在我的Python项目中，有很多代码如下所示：

# the codes I with to simplify
for _ in range(3):
    try:
        # do something
        break
    except:
        print "Exception. Retrying..."
        continue
 else:
    print "ERROR!"

我知道我可以使用contextmanager简化“尝试…除了…”部分，方法如下：

# define the exception handler function first
@contextmanager
def ignore_exception():
    try:
        yield
    except:
        print "Exception. Retrying..."
        continue

# then use "with" to simplify the codes
with ignore_exception:
    # do something

但是，我不能以相同的方式添加“for”循环，即

# define the exception handler function with for loop first
@contextmanager
def ignore_exception():
    try:
        for _ in range(3):
            yield
            break
        else:
            print "ERROR!"
    except:
        print "Exception. Retrying..."
        continue

# then use "with" to simplify the codes
with ignore_exception:
    # do something

将引发一个异常：

/usr/lib/python2.7/contextlib.pyc in __exit__(self, type, value, traceback)
     26                 return
     27             else:
---> 28                 raise RuntimeError("generator didn't stop")
     29         else:
     30             if value is None:

RuntimeError: generator didn't stop

有没有办法克服这种错误并将我的代码简化为：

# ideal codes
with xxxxxx: # don't have to use "with" statement
    # do something

---

只需编写一个decorator，它可以处理循环和异常忽略部分，如下所示

def trys(maximum_tries=3):
    def decorator(func):
        def inner(*args, **kwargs):
            for _ in range(1, maximum_tries + 1):
                try:
                    return func(*args, **kwargs)
                except ArithmeticError, e:
                    print("Error : [{}], Retrying Attempt {}...".format(e, _))
                else:
                    break
        return inner
    return decorator

outer_a = 0


@trys(maximum_tries=4)
def adder(a):
    global outer_a
    outer_a += 1
    if outer_a < 4:
        raise ArithmeticError("Cannot add with {}".format(outer_a))
    else:
        return a + outer_a

print(adder(0))

现在，您可以这样调用它

def trys(maximum_tries=3):
    def decorator(func):
        def inner(*args, **kwargs):
            for _ in range(1, maximum_tries + 1):
                try:
                    return func(*args, **kwargs)
                except ArithmeticError, e:
                    print("Error : [{}], Retrying Attempt {}...".format(e, _))
                else:
                    break
        return inner
    return decorator

outer_a = 0


@trys(maximum_tries=4)
def adder(a):
    global outer_a
    outer_a += 1
    if outer_a < 4:
        raise ArithmeticError("Cannot add with {}".format(outer_a))
    else:
        return a + outer_a

print(adder(0))

---

注意：有一个名为的开源库，它以更好、更灵活的方式完成相同的任务。如果可能，您可能希望检查并使用它，而不是滚动您自己的装饰程序。不要重复你自己：-）

这意味着我必须将所有内容都放在函数中才能使用装饰器。有没有一种方法可以避免像宏在C/C++中所能做的那样使用函数。例如，我只想重试第n到n+10行3次，然后重试第n+11到n+20行2次。为什么contextmanager不允许在内部使用循环？