How can I open multiple files using "with open" in Python?


Question

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?

1
587
12/29/2014 8:07:23 AM

Accepted Answer

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.


In the rare case that you want to open a variable number of files all at the same time, you can use contextlib.ExitStack, starting from Python version 3.3:

with ExitStack() as stack:
    files = [stack.enter_context(open(fname)) for fname in filenames]
    # Do something with "files"

Most of the time you have a variable set of files, you likely want to open them one after the other, though.

911
2/10/2019 2:04:46 PM

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

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