Is there a more pythonic way to open a file if given one as an argument or stdin if not?

Question

I'm trying to write a python script which follows the common unix command line pattern of accepting input from stdin if no file name is given. This is what I've been using:

if __name__ == "__main__":
    if len(sys.argv) > 1:
        stream = open(sys.argv[1])
    else:
        stream = sys.stdin

Is there a more pythonic way to do that?

Answer 1

similar but one-line solution

stream = sys.argv[1] if len(sys.argv)>1 else sys.stdin

Answer 2

The fileinput module is perfect for this.

Answer 3

I would suggest you make it more unixy instead:

if len(sys.argv) > 1:
    sys.stdin = open(sys.argv[1])

Answer 4

how about this one?

stream=sys.argv[1:] and open(sys.argv[1]) or sys.stdin