SystemExit: 2 error when calling parse_args() within ipython

Question

I'm learning basics of Python and got already stuck at the beginning of argparse tutorial. I'm getting the following error:

import argparse
parser = argparse.ArgumentParser()
args = parser.parse_args()

usage: __main__.py [-h] echo
__main__.py: error: unrecognized arguments: -f
An exception has occurred, use %tb to see the full traceback.

SystemExit: 2

a %tb command gives the following output:

SystemExit                                Traceback (most recent call last)
<ipython-input-16-843cc484f12f> in <module>()
----> 1 args = parser.parse_args()

C:\Users\Haik\Anaconda2\lib\argparse.pyc in parse_args(self, args, namespace)
   1702         if argv:
   1703             msg = _('unrecognized arguments: %s')
-> 1704             self.error(msg % ' '.join(argv))
   1705         return args
   1706 

C:\Users\Haik\Anaconda2\lib\argparse.pyc in error(self, message)
   2372         """
   2373         self.print_usage(_sys.stderr)
-> 2374         self.exit(2, _('%s: error: %s\n') % (self.prog, message))

C:\Users\Haik\Anaconda2\lib\argparse.pyc in exit(self, status, message)
   2360         if message:
   2361             self._print_message(message, _sys.stderr)
-> 2362         _sys.exit(status)
   2363 
   2364     def error(self, message):

SystemExit: 2

How could I fix this problem?

Condign

You cannot experiment with this easily in ipython as the ipython command line will be used per default in parse_args. Try this with a normal *.py file and the python executable to launch that file. – Bellyband

See this argparse. It will help to you or surely helps to others – Supinate

I've reproduced your code and I don't find any problem. Have you tried it in .py file instead of another way? – Delate

Alternatively in ipython you can pass a list of args to the parser. From the argparse documentation: parser.parse_args(['--sum', '7', '-1', '42']) or parser.parse_args('--sum 7 -1 42'.split() ) – Interspace

thank you guys for immediate help pointing out where my mistake was and also providing alternative way, I've tried in different ways but all with ipython and not "normal" *.py – Condign

Please check out my answer below. I spent 3 hours on this and I think I found the real solution (and it should be the accepted answer to help people with the same problem). I took my answer from here #48796669 – Quinquagesima

Answer 1

argparse is a module designed to parse the arguments passed from the command line, so for example if you type the following at a command prompt:

$ python my_programme.py --arg1=5 --arg2=7

You can use argparse to interpret the --arg1=5 --arg2=7 part. If argparse thinks the arguments are invalid, it exits, which in general is done in python by calling sys.exit() which raises the SystemExit error, which is what you're seeing.

So the problem is you're trying to use argparse from an interactive interpreter (looks like ipython), and at this point the programme has already started, so the args should have already been parsed.

To try it properly create a separate python file such as my_programme.py and run it using python from a command line, as I illustrated.

Avellaneda

yes it was exactly ipython, just want to say how much i appreciate your clear answer! And even I've spent several hours trying to figure out it myself, it seems I would need some more days to solve this problem alone. Thank you. – Condign

Answer 2

[quick solution] Add an dummy parser argument in the code

parser.add_argument('-f')

Keloid

Hey, it worked but can you please provide some explanation on how does it working? – Schoolroom

Jupyter-Notebook has some some default arguments set. One of which is "-f ". import sys sys.argv running the above code will display them – Keloid

This should be the accepted answer in my opinion. The accepted answer is not a solution, my code is a Jupyter notebook I cannot run it as a script in the command line. – Quinquagesima

Answer 3

had run into a similar issue. adding these lines fixed the issue for me.

import sys
sys.argv=['']
del sys

Horsehide

Umm... why did that help? Can you explain this for anyone that comes reading it? – Blowing

Helpful answer, similar to https://mcmap.net/q/325386/-how-to-call-module-written-with-argparse-in-ipython-notebook – Pandemic

Answer 4

I'm surprised nobody mentioned this answer here How to fix ipykernel_launcher.py: error: unrecognized arguments in jupyter?

There is no need for the -f argument. Also, the -f trick works for Jupyter but not in VS code.

tl;dr

Use

args, unknown = parser.parse_known_args()

INSTEAD of

args = parser.parse_args()

Quinquagesima

Answer 5

parse_args method, when it's called without arguments, attempts to parse content of sys.argv. Your interpreter process had filled sys.argv with values that does not match with arguments supported by your parser instance, that's why parsing fails.

Try printing sys.argv to check what arguments was passed to your interpreter process.

Try calling parser.parse_args(['my', 'list', 'of', 'strings']) to see how parser will work for interpreter launched with different cmdline arguments.

Zamir

Answer 6

There are two ways of solving this:

1. Use get_ipython().__class__.__name__ to determine whether we're running on ipython or terminal, and then use parser.parse_args(args = []) when we're running on ipython

   try:
       get_ipython().__class__.__name__
       # No error means we're running on ipython
       args = parser.parse_args(args = []) # Reset args
   except NameError:
       # NameError means that we're running on terminal
       args = parser.parse_args()
   

2. Use parser.parse_known_args() to store existing args separately. We would get a return of a tuple with two values (first is args that are added by add_argument and second is existing args)

   args = parser.parse_known_args()[0] # Allow unrecognized arguments
   

The difference of these two approaches is that the second one will allow unrecognized arguments. It will be stored in the second value of the returned tuple.

Exist

I find the second approach minimal invasive, thank you very much! – Evaporimeter

Answer 7

I know this question is nearly three years old but as dumb as it can sound, this exit error is also produced when you don't have argparse installed instead of the default "This module can't be found" error message. Just helping people that may have this error aswell.

Sevilla

How can it be not installed? This is a standard library module. – Bonnee

Answer 8

Add an argument and assign some value works. I was passing args (ArgumentParser type object) from one function to another (not in a typical case, like, getting user args from terminal).

from argparse import ArgumentParser
parser = ArgumentParser()
# create and assign a dummy args
parser.add_argument('--myarg1')
args = parser.parse_args(['--myarg1', ''])

args.myarg2 = True  # actuals args assignment
myTargetFunction(args)  # passing args to another function  

I found without any actual args in parser, parse_args() gives error.

Barometer