关于python：’with’语句中有多个变量？


Multiple variables in a 'with' statement?

是否可以在Python中使用with语句声明多个变量？

就像是：

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from __future__ import with_statement

with open("out.txt","wt"), open("in.txt") as file_out, file_in:
    for line in file_in:
        file_out.write(line)

......还是正在清理两个资源同时出现问题？

相关讨论

从v3.1和Python 2.7开始，它可以在Python 3中实现。新的with语法支持多个上下文管理器：

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with A() as a, B() as b, C() as c:
    doSomething(a,b,c)

与contextlib.nested不同，这保证即使C()或__enter__()方法引发异常，a和b也会调用__exit__()。

相关讨论

contextlib.nested支持这个：

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import contextlib

with contextlib.nested(open("out.txt","wt"), open("in.txt")) as (file_out, file_in):

   ...

更新：
引用文档，关于contextlib.nested：

Deprecated since version 2.7: The with-statement now supports this
functionality directly (without the confusing error prone quirks).

看拉法？ Dowgird的回答是为了获取更多信息。

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请注意，如果将变量拆分为行，则必须使用反斜杠来换行换行符。

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with A() as a, \
     B() as b, \
     C() as c:
    doSomething(a,b,c)

括号不起作用，因为Python会创建一个元组。

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with (A(),
      B(),
      C()):
    doSomething(a,b,c)

由于元组缺少__enter__属性，因此会出现错误(无法识别并且不识别类类型)：

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AttributeError: __enter__

如果您尝试在括号内使用as，Python会在解析时捕获错误：

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with (A() as a,
      B() as b,
      C() as c):
    doSomething(a,b,c)

SyntaxError: invalid syntax

https://bugs.python.org/issue12782似乎与此问题有关。

我想你想这样做：

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from __future__ import with_statement

with open("out.txt","wt") as file_out:
    with open("in.txt") as file_in:
        for line in file_in:
            file_out.write(line)
相关讨论

从Python 3.3开始，您可以使用contextlib模块中的类ExitStack。

它可以管理动态数量的上下文感知对象，这意味着如果您不知道要处理多少文件，它将特别有用。

文档中提到的规范用例是管理动态数量的文件。

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with ExitStack() as stack:
    files = [stack.enter_context(open(fname)) for fname in filenames]
    # All opened files will automatically be closed at the end of
    # the with statement, even if attempts to open files later
    # in the list raise an exception

这是一个通用的例子：

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from contextlib import ExitStack

class X:
    num = 1

    def __init__(self):
        self.num = X.num
        X.num += 1

    def __repr__(self):
        cls = type(self)
        return '{cls.__name__}{self.num}'.format(cls=cls, self=self)

    def __enter__(self):
        print('enter {!r}'.format(self))
        return self.num

    def __exit__(self, exc_type, exc_value, traceback):
        print('exit {!r}'.format(self))
        return True

xs = [X() for _ in range(3)]

with ExitStack() as stack:
    print(stack._exit_callbacks)
    nums = [stack.enter_context(x) for x in xs]
    print(stack._exit_callbacks)
print(stack._exit_callbacks)
print(nums)

输出：

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deque([])
enter X1
enter X2
enter X3
deque([<function ExitStack._push_cm_exit.<locals>._exit_wrapper at 0x7f5c95f86158>, <function ExitStack._push_cm_exit.<locals>._exit_wrapper at 0x7f5c95f861e0>, <function ExitStack._push_cm_exit.<locals>._exit_wrapper at 0x7f5c95f86268>])
exit X3
exit X2
exit X1
deque([])
[1, 2, 3]