是否可以在python中使用`with`打开任意数量的项目？

Question

我有一个情况,我有几个项目,我想用with块打开.在我的情况下,这些是外部硬件设备,在关闭时需要进行一些清理 - 但这对于手头的要点并不重要.

假设一个类是这样的:

class Controller(object):

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

    def __enter__(self):
        # Do some work on entry
        print("Entering", self._name)
        return self

    def __exit__(self, type, value, traceback):
        # Clean up (restoring external state, turning off hardware, etc)
        print("Exiting", self._name)
        return False

    def work(self):
        print("Working on", self._name)

我会(给定一定数量的Controllers),做类似的事情

with Controller("thing1") as c1:
    with Controller("thing2") as c2:
        c1.do_work()
        c2.do_work()

但是,我遇到过这样一种情况:我需要以这种方式管理一些灵活的事情.也就是说,我的情况类似于:

things = ["thing1", "thing2", "thing3"] # flexible in size
for thing in things:
    with Controller(thing) as c:
        c.do_work()

但是,上面并没有完全按照我的需要做 - 这是同时拥有范围内的Controllers所有内容thing.

我已经构建了一个通过递归工作的玩具示例:

def with_all(controllers, f, opened=None):
    if opened is None:
        opened = []

    if controllers:
        with controllers[0] as t:
            opened.append(t)
            controllers = controllers[1:]

            with_all(controllers, f, opened)
    else:
        f(opened)

def do_work_on_all(controllers):
    for c in controllers:
        c.work()

names = ["thing1", "thing2", "thing3"]
controllers = [Controller(n) for n in names]

with_all(controllers, do_work_on_all)

但我不喜欢实际函数调用的递归或抽象.我对以更"pythonic"方式做这件事的想法感兴趣.

Answer 1

是的,有更多的pythonic方法,使用标准库contextlib,它有一个类ExitStack,它可以做你想要的东西:

with ExitStack() as stack:
    controllers = [stack.enter_context(Controller(n)) for n in names]

这应该做你想要的.