CODEWITHSUNDEEP
pythonpython-2.7
anita
anita

Reputation: 177

Open list of files using a with statement

I'm trying to open several files from sys.argv inputs using a with statement.

I know I can do it by manually typing out each one:

with open(sys.argv[1], 'r') as test1, open(sys.argv[2], 'r') as test2, \
     open(sys.argv[3], 'r') as test3, open(sys.argv[4], 'r') as test4:
    do_something()

But is there a way to do this without doing so, like the following pseudocode:

with open(sys.argv[1:4], 'r') as test1, test2, test3:
    do_something()

Upvotes: 3

Views: 635

Answers (1)

Eric
Eric

Reputation: 97641

You can do this in Python 3.3+ with contextlib.ExitStack:

from contextlib import ExitStack

with ExitStack() as stack:
    files = [stack.enter_context(open(arg, 'r')) for arg in sys.arv[1:4]]

Amusingly, the example in the documentation is exactly what you want.

This correctly closes any open files upon exiting the with statement - even if something went wrong before they were all open

For earlier version of python, there is a backport in the contextlib2 package, which you can get through pip

Upvotes: 2

Related Questions