Open a list of files using with/as context manager

Question

Note: I am aware of the

with open('f1') as f1, open('f2') as f2:     ... 

syntax. This is a different question.

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Given a list of strings file_names is there a way using with/as to open every file name in that using a single line. Something such as:

with [open(fn) for fn in file_names] as files:     # use the list of files 

which of course doesn't work as it attempts to use the context manager on a list. The length of the list may not be known until run-time, such as sys.argv[1:]

Answer 1

If you have access to Python 3.3+, there is a special class designed exactly for this purpose: the ExitStack. It works just like you'd expect:

with contextlib.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 

Answer 2

How about this?

class ListContext:     def __init__(self, l):         self.l = l      def __enter__(self):         for x in self.l:             x.__enter__()         return self.l      def __exit__(self, type, value, traceback):         for x in self.l:             x.__exit__(type, value, traceback)  arr = ['a', 'b', 'c']  with ListContext([open(fn, 'w') for fn in arr]) as files:     print files  print files 

Output is:

[<open file 'a', mode 'w' at 0x7f43d655e390>, <open file 'b', mode 'w' at 0x7f43d655e420>, <open file 'c', mode 'w' at 0x7f43d655e4b0>] [<closed file 'a', mode 'w' at 0x7f43d655e390>, <closed file 'b', mode 'w' at 0x7f43d655e420>, <closed file 'c', mode 'w' at 0x7f43d655e4b0>] 

Notice, they are open inside the with context and closed outside.

This is using the Python context manager API.

EDIT: It seems like this already exists but is deprecated: See contextlib and this SO question. Use it like this:

import contextlib  with contextlib.nested(*[open(fn, 'w') for fn in arr]) as files:     print files print files