python - 是否可以重试一段特定的代码而不考虑异常？

Question

在我的Python项目中，像下面这样的代码非常多:

# 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 来简化“try ... except ...”部分，方法是:

# 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

有没有办法避免这样的错误并将我的代码简化为:

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

Answer 1

只需要写一个装饰器，它可以同时处理循环和异常忽略部分，就像这样

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

现在，您可以像这样调用它

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))

它不断尝试将当前值添加到 outer_a 并且只有当它大于或等于 4 时才可以添加到它。因此，它不断重试直到最大尝试次数筋疲力尽。它捕获抛出的异常，将其打印到标准输出并重试。所以，输出将是

Error : [Cannot add with 1], Retrying Attempt 1...
Error : [Cannot add with 2], Retrying Attempt 2...
Error : [Cannot add with 3], Retrying Attempt 3...
4

注意:有一个名为retrying的开源库它以更好和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器。不要重复自己:-)