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在Python中调用上下文管理器的多种方法

contextmanager python-2.x class python-2.7 python

Cloud · 技术社区 · 7 年前

背景

我在Python中有一个类,它接受互斥体列表。然后对该列表进行排序,并使用 __enter__() 和 __exit__() 以特定顺序锁定/解锁所有互斥锁,以防止死锁。

类目前为我们节省了许多潜在死锁的麻烦,因为我们可以在 RAII style ,即:

self.lock = SuperLock(list_of_locks)
# Lock all mutexes.
with self.lock:
    # Issue calls to all hardware protected by these locks.

问题

我们希望公开这个类提供一个raii风格的api的方法,这样当以某种方式调用时,我们就可以一次锁定一半的互斥体,即:

self.lock = SuperLock(list_of_locks)
# Lock all mutexes.
with self.lock:
    # Issue calls to all hardware protected by these locks.

# Lock the first half of the mutexes in SuperLock.list_of_locks
with self.lock.first_half_only:
    # Issue calls to all hardware protected by these locks.

# Lock the second half of the mutexes in SuperLock.list_of_locks
with self.lock.second_half_only:
    # Issue calls to all hardware protected by these locks.

问题

是否有方法提供这种类型的功能以便我可以调用 with self.lock.first_half_only 或 with self.lock_first_half_only () 为用户提供一个简单的API?我们希望将所有这些功能保存在一个类中。

谢谢您。

3 回复 | 直到 7 年前

wim 7 年前

是的,您可以得到这个接口。将在WITH语句上下文中输入/退出的对象是已解析的属性。因此,您可以继续将上下文管理器定义为上下文管理器的属性:

from contextlib import ExitStack  # pip install contextlib2
from contextlib import contextmanager

@contextmanager
def lock(name):
    print("entering lock {}".format(name))
    yield
    print("exiting lock {}".format(name))

@contextmanager
def many(contexts):
    with ExitStack() as stack:
        for cm in contexts:
            stack.enter_context(cm)
        yield

class SuperLock(object):

    def __init__(self, list_of_locks):
        self.list_of_locks = list_of_locks

    def __enter__(self):
        # implement for entering the `with self.lock:` use case
        return self

    def __exit__(self, exce_type, exc_value, traceback):
        pass

    @property
    def first_half_only(self):
        return many(self.list_of_locks[:4])

    @property
    def second_half_only(self):
        # yo dawg, we herd you like with-statements
        return many(self.list_of_locks[4:])

当您创建并返回一个新的上下文管理器时,您可以使用实例中的状态(即 self )

示例用法:

>>> list_of_locks = [lock(i) for i in range(8)] 
>>> super_lock = SuperLock(list_of_locks) 
>>> with super_lock.first_half_only: 
...     print('indented') 
...   
entering lock 0
entering lock 1
entering lock 2
entering lock 3
indented
exiting lock 3
exiting lock 2
exiting lock 1
exiting lock 0

编辑 :基于类的等价于 lock 上面显示的生成器上下文管理器

class lock(object):

    def __init__(self, name):
        self.name = name

    def __enter__(self):
        print("entering lock {}".format(self.name))
        return self

    def __exit__(self, exce_type, exc_value, traceback):
        print("exiting lock {}".format(self.name))
        # If you want to handle the exception (if any), you may use the
        # return value of this method to suppress re-raising error on exit

Sraw 7 年前

from contextlib import contextmanager

class A:

    @contextmanager
    def i_am_lock(self):
        print("entering")
        yield
        print("leaving")

a = A()

with a.i_am_lock():
    print("inside")

输出:

entering
inside
leaving

您还可以使用 contextlib.ExitStack 更好地管理你的锁。

Felix 7 年前

我会用一个 SimpleNamespace 允许属性访问不同的 SuperLock 对象,例如:

from types import SimpleNamespace

self.lock = SimpleNamespace(
    all=SuperLock(list_of_locks),
    first_two_locks=SuperLock(list_of_locks[:2]),
    other_locks=SuperLock(list_of_locks[2:])
)

with self.lock.all:
    # Issue calls to all hardware protected by these locks.

with self.lock.first_two_locks:
    # Issue calls to all hardware protected by these locks.

with self.lock.other_locks:
    # Issue calls to all hardware protected by these locks.

编辑:

对于Python2,可以使用这个类来实现类似的行为:

class SimpleNamespace:
    def __init__(self, **kwargs):
        self.__dict__.update(kwargs)