How can I open multiple files using “with open” in

问题:

I want to change a couple of files at one time, iff I can write to all of them. I\'m wondering if I somehow can combine the multiple open calls with the with statement:

try:
  with open(\'a\', \'w\') as a and open(\'b\', \'w\') as b:
    do_something()
except IOError as e:
  print \'Operation failed: %s\' % e.strerror

If that\'s not possible, what would an elegant solution to this problem look like?

回答1:

As of Python 2.7 (or 3.1 respectively) you can write

with open(\'a\', \'w\') as a, open(\'b\', \'w\') as b:
    do_something()

In earlier versions of Python, you can sometimes use contextlib.nested() to nest context managers. This won\'t work as expected for opening multiples files, though -- see the linked documentation for details.

回答2:

Just replace and with , and you\'re done:

try:
    with open(\'a\', \'w\') as a, open(\'b\', \'w\') as b:
        do_something()
except IOError as e:
    print \'Operation failed: %s\' % e.strerror

回答3:

For opening many files at once or for long file paths, it may be useful to break things up over multiple lines. From the Python Style Guide as suggested by @Sven Marnach in comments to another answer:

with open(\'/path/to/InFile.ext\', \'r\') as file_1, \\
     open(\'/path/to/OutFile.ext\', \'w\') as file_2:
    file_2.write(file_1.read())

回答4:

Nested with statements will do the same job, and in my opinion, are more straightforward to deal with.

Let\'s say you have inFile.txt, and want to write it into two outFile\'s simultaneously.

with open(\"inFile.txt\", \'r\') as fr:
    with open(\"outFile1.txt\", \'w\') as fw1:
        with open(\"outFile2.txt\", \'w\') as fw2:
            for line in fr.readlines():
                fw1.writelines(line)
                fw2.writelines(line)

EDIT:

I don\'t understand the reason of the downvote. I tested my code before publishing my answer, and it works as desired: It writes to all of outFile\'s, just as the question asks. No duplicate writing or failing to write. So I am really curious to know why my answer is considered to be wrong, suboptimal or anything like that.

回答5:

With python 2.6 It will not work, we have to use below way to open multiple files:

with open(\'a\', \'w\') as a:
    with open(\'b\', \'w\') as b:

回答6:

Since Python 3.3, you can use the class ExitStack from the contextlib module to safely
open an arbitrary number of files.

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.

In fact, 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

---

If you are interested in the details, here is a generic example in order to explain how ExitStack operates:

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(len(stack._exit_callbacks)) # number of callbacks called on exit
    nums = [stack.enter_context(x) for x in xs]
    print(len(stack._exit_callbacks))

print(len(stack._exit_callbacks))
print(nums)

Output:

0
enter X1
enter X2
enter X3
3
exit X3
exit X2
exit X1
0
[1, 2, 3]