Python 3.x call function with sys.argv[]

Question

I have a function that processes file contents, but right now I have the filename hardcoded in the function like this as a keyword argument:

def myFirstFunc(filename=open('myNotes.txt', 'r')): 
    pass

and I call it like this:

myFirstFunc()

I would like to treat the argument as a filename and process the contents.

1. How do I modify the statement above? I tried this:

   filename=sys.argv[1]  # or is it 0?
   

2. How do I call it?

Answer 1

something like this:

#!/usr/bin/python3

import sys


def myFirstFunction():
    return open(sys.argv[1], 'r')

openFile = myFirstFunction()

for line in openFile:
    print (line.strip()) #remove '\n'? if not remove .strip()
    #do other stuff

openFile.close() #don't forget to close open file

then I would call it like the following:

./readFile.py temp.txt

which would output the contents of temp.txt

sys.argv[0] outputs the name of script. In this case ./readFile.py

Updating My Answer
because it seems others want a try approach

How do I check whether a file exists using Python? is a good question on this subject of how to check if a file exists. There appears to be a disagreement on which method to use, but using the accepted version it would be as followed:

 #!/usr/bin/python3

import sys


def myFirstFunction():
    try:
        inputFile = open(sys.argv[1], 'r')
        return inputFile
    except Exception as e:
        print('Oh No! => %s' %e)
        sys.exit(2) #Unix programs generally use 2 for 
                    #command line syntax errors
                    # and 1 for all other kind of errors.


openFile = myFirstFunction()

for line in openFile:
    print (line.strip())
    #do other stuff
openFile.close()

which would output the following:

$ ./readFile.py badFile
Oh No! => [Errno 2] No such file or directory: 'badFile'

you could probably do this with an if statement, but I like this comment on EAFP VS LBYL