Multiple variables in a 'with' statement?

Question

Is it possible to declare more than one variable using a with statement in Python?

Something like:

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)

... or is cleaning up two resources at the same time the problem?

Answer 1

It is possible in Python 3 since v3.1 and Python 2.7. The new with syntax supports multiple context managers:

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

Unlike the contextlib.nested, this guarantees that a and b will have their __exit__()'s called even if C() or it's __enter__() method raises an exception.

Answer 2

contextlib.nested supports this:

import contextlib

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

   ...

Update:
To quote the documentation, regarding contextlib.nested:

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

See Rafał Dowgird's answer for more information.

Answer 3

Note that if you split the variables into lines, you must use backslashes to wrap the newlines.

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

Parentheses don't work, since Python creates a tuple instead.

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

Since tuples lack a __enter__ attribute, you get an error (undescriptive and does not identify class type):

AttributeError: __enter__

If you try to use as within parentheses, Python catches the mistake at parse time:

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

SyntaxError: invalid syntax

https://bugs.python.org/issue12782 seems to be related to this issue.

Answer 4

I think you want to do this instead:

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)

Answer 5

Since Python 3.3, you can use the class ExitStack from the contextlib module.

It can manage a dynamic number of context-aware objects, which means that it will prove especially useful if you don't know how many files you are going to handle.

The canonical use-case that is mentioned in the documentation is managing a dynamic number of files.

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

Here is a generic example:

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)

Output:

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]