Arguments get mixed up with argparse - python

def main(argv):
parser = argparse.ArgumentParser()
parser.add_argument('p', 'projectType', type=str, help = 'c or c++ project type', choices=['c', 'c++'])
parser.add_argument('i', 'inputfile', type=pathlib.Path, help = 'the input file path')
parser.add_argument('o',"outputfile",type=pathlib.Path, help= 'the output file path')
parser.add_argument
args = parser.parse_args()
when i run this code on command prompt, and enter the inputfile first for example.. it gives me an error. does Argparse have an option like getopt (where i call something with a letter before inputting it so no mix up ex, '-i ...input...'). Here this option is just for optional arguments.
Is there a way for positional arguments to do that?

As per the argparse documentation you would define it as an "optional" argument but with the required flag set to true, e.g.:
parser.add_argument('-i', '--inputfile', required=True, type=pathlib.Path, help='the input file path')

Related

python argparse default with nargs wont work [duplicate]

This question already has answers here:
Argparse optional argument with different default if specified without a value
(2 answers)
Closed last month.
Here is my code:
from argparse import ArgumentParser, RawTextHelpFormatter
example_text = "test"
parser = ArgumentParser(description='my script.',
epilog=example_text,
formatter_class=RawTextHelpFormatter)
parser.add_argument('host', type=str, default="10.10.10.10",
help="Device IP address or Hostname.")
parser.add_argument('-j','--json_output', type=str, default="s", nargs='?',choices=["s", "l"],
help="Print GET statement in json form.")
#mutally exclusive required settings supplying the key
settingsgroup = parser.add_mutually_exclusive_group(required=True)
settingsgroup.add_argument('-k', '--key', type=str,
help="the api-key to use. WARNING take care when using this, the key specified will be in the user's history.")
settingsgroup.add_argument('--config', type=str,
help="yaml config file. All parameters can be placed in the yaml file. Parameters provided from form command line will take priority.")
args = parser.parse_args()
print(args.json_output)
my output:
None
Everything I am reading online says this should work, but it doesn't. Why?
You could use the const= parameter. const stores its value when the option is present but have no values
import argparse
parser = argparse.ArgumentParser()
parser.add_argument('-j', '--json-output', nargs='?', choices=['s', 'l'], const='s')
args = parser.parse_args()
However design wise, it might be better to use the following:
import argparse
parser = argparse.ArgumentParser()
parser.add_argument('-o', '--output-type', choices=['json-s', 'json-l', 'normal'], default='normal')
args = parser.parse_args()

Argparse Unable to access optional argument with specific name [duplicate]

I want to have some options in argparse module such as --pm-export however when I try to use it like args.pm-export I get the error that there is not attribute pm. How can I get around this issue? Is it possible to have - in command line options?
As indicated in the argparse docs:
For optional argument actions, the value of dest is normally inferred from the option strings. ArgumentParser generates the value of dest by taking the first long option string and stripping away the initial -- string. Any internal - characters will be converted to _ characters to make sure the string is a valid attribute name
So you should be using args.pm_export.
Unfortunately, dash-to-underscore replacement doesn't work for positional arguments (not prefixed by --).
E.g:
parser = argparse.ArgumentParser(description='Process some integers.')
parser.add_argument('logs-dir',
help='Directory with .log and .log.gz files')
parser.add_argument('results-csv', type=argparse.FileType('w'),
default=sys.stdout,
help='Output .csv filename')
args = parser.parse_args()
print args
# gives
# Namespace(logs-dir='./', results-csv=<open file 'lool.csv', mode 'w' at 0x9020650>)
So, you should use 1'st argument to add_argument() as attribute name and metavar kwarg to set how it should look in help:
parser = argparse.ArgumentParser(description='Process some integers.')
parser.add_argument('logs_dir', metavar='logs-dir',
nargs=1,
help='Directory with .log and .log.gz files')
parser.add_argument('results_csv', metavar='results-csv',
nargs=1,
type=argparse.FileType('w'),
default=sys.stdout,
help='Output .csv filename')
args = parser.parse_args()
print args
# gives
# Namespace(logs_dir=['./'], results_csv=[<open file 'lool.csv', mode 'w' at 0xb71385f8>])
Dashes are converted to underscores:
import argparse
pa = argparse.ArgumentParser()
pa.add_argument('--foo-bar')
args = pa.parse_args(['--foo-bar', '24'])
print args # Namespace(foo_bar='24')
Concise and explicit but probably not always acceptable way would be to use vars():
#!/usr/bin/env python3
import argparse
parser = argparse.ArgumentParser()
parser.add_argument('a-b')
args = vars(parser.parse_args())
print(args['a-b'])
getattr(args, 'positional-arg')
This is another OK workaround for positional arguments:
#!/usr/bin/env python3
import argparse
parser = argparse.ArgumentParser()
parser.add_argument('a-b')
args = parser.parse_args(['123'])
assert getattr(args, 'a-b') == '123'
Tested on Python 3.8.2.
I guess the last option is to change shorten option -a to --a
import argparse
parser = argparse.ArgumentParser(description="Help")
parser.add_argument("--a", "--argument-option", metavar="", help="") # change here
args = parser.parse_args()
option = args.a # And here
print(option)

Python calling a module that uses argparser

This is probably a silly question, but I have a python script that current takes in a bunch of arguments using argparser and I would like to load this script as a module in another python script, which is fine. But I am not sure how to call the module as no function is defined; can I still call it the same way I do if I was just invoking it from cmd?
Here is the child script:
import argparse as ap
from subprocess import Popen, PIPE
parser = ap.ArgumentParser(
description='Gathers parameters.')
parser.add_argument('-f', metavar='--file', type=ap.FileType('r'), action='store', dest='file',
required=True, help='Path to json parameter file')
parser.add_argument('-t', metavar='--type', type=str, action='store', dest='type',
required=True, help='Type of parameter file.')
parser.add_argument('-g', metavar='--group', type=str, action='store', dest='group',
required=False, help='Group to apply parameters to')
# Gather the provided arguments as an array.
args = parser.parse_args()
... Do stuff in the script
and here is the parent script that I want to invoke the child script from; it also uses arg parser and does some other logic
from configuration import parameterscript as paramscript
# Can I do something like this?
paramscript('parameters/test.params.json', test)
Inside the configuration directory, I also created an init.py file that is empty.
The first argument to parse_args is a list of arguments. By default it's None which means use sys.argv. So you can arrange your script like this:
import argparse as ap
def main(raw_args=None):
parser = ap.ArgumentParser(
description='Gathers parameters.')
parser.add_argument('-f', metavar='--file', type=ap.FileType('r'), action='store', dest='file',
required=True, help='Path to json parameter file')
parser.add_argument('-t', metavar='--type', type=str, action='store', dest='type',
required=True, help='Type of parameter file.')
parser.add_argument('-g', metavar='--group', type=str, action='store', dest='group',
required=False, help='Group to apply parameters to')
# Gather the provided arguments as an array.
args = parser.parse_args(raw_args)
print(vars(args))
# Run with command line arguments precisely when called directly
# (rather than when imported)
if __name__ == '__main__':
main()
And then elsewhere:
from first_module import main
main(['-f', '/etc/hosts', '-t', 'json'])
Output:
{'group': None, 'file': <_io.TextIOWrapper name='/etc/hosts' mode='r' encoding='UTF-8'>, 'type': 'json'}
There may be a simpler and more pythonic way to do this, but here is one possibility using the subprocess module:
Example:
child_script.py
import argparse
parser = argparse.ArgumentParser()
parser.add_argument("-n", "--name", help="your name")
args = parser.parse_args()
print("hello there {}").format(args.name)
Then another Python script can call that script like so:
calling_script.py:
import subprocess
# using Popen may suit better here depending on how you want to deal
# with the output of the child_script.
subprocess.call(["python", "child_script.py", "-n", "Donny"])
Executing the above script would give the following output:
"hello there Donny"
One of the option is to call it as subprocess call like below:
import subprocess
childproc = subprocess.Popen('python childscript.py -file yourjsonfile')
op, oe = childproc.communicate()
print op

Unable to assign default values to arguments in Python argparse

I am using Python argparse to take in parameters through the CLI. I have tried using the following but when I don't give any one of the arguments, it gives the output as None. I want then default ones to be the ones provided in const=. Please take a look.
parser = argparse.ArgumentParser()
parser.add_argument('--input', nargs='?', const='testInput')
parser.add_argument('--target', nargs='?', const='testTarget')
parser.add_argument('--msg', nargs='?', const='helloFromTheOtherSide')
args = parser.parse_args()
print args.input
If I don't give input, it prints it as None as I said. I want it to print TestInput instead..
Use the default argument:
parser = argparse.ArgumentParser()
parser.add_argument('--input', nargs='?', default='testInput')
parser.add_argument('--target', nargs='?', default='testTarget')
parser.add_argument('--msg', nargs='?', default='helloFromTheOtherSide')
args = parser.parse_args()
print args.input
With
parser.add_argument('--input', nargs='?', default='testInput', const='aConst')
you have a 3 way choice
prog # input='testInput'
prog --input # input='aConst'
prog --input myfile # input='myfile'
If you don't need that aConst option, omit the nargs='?'. Since it is a flagged argument it is already optional. It doesn't need the `?'.
parser.add_argument('--input', default='testInput')

argparse coding issue

write a script that takes two optional boolean arguments,"--verbose‚" and ‚"--live", and two required string arguments, "base"and "pattern". Please set up the command line processing using argparse.
This is the code I have so far for the question, I know I am getting close but something is not quite right. Any help is much appreciated.Thanks for all the quick useful feedback.
def main():
import argparse
parser = argparse.ArgumentParser(description='')
parser.add_argument('base', type=str)
parser.add_arguemnt('--verbose', action='store_true')
parser.add_argument('pattern', type=str)
parser.add_arguemnt('--live', action='store_true')
args = parser.parse_args()
print(args.base(args.pattern))
The string arguments are not required by default, so you need to state that. Also the print statement that uses the arguments is incorrect.
#!/usr/bin/python
import argparse
if __name__=="__main__":
parser = argparse.ArgumentParser(description='eg $python myargs.py --base arg1 --pattern arg2 [--verbose] [--live]')
parser.add_argument('--base', required=True, type=str)
parser.add_argument('--pattern', required=True, type=str)
parser.add_argument('--verbose', action='store_true')
parser.add_argument('--live', action='store_true')
args = parser.parse_args()
print "args.base=" + str(args.base)
print "args.pattern=" + str(args.pattern)
print "args.verbose=" + str(args.verbose)
print "args.live=" + str(args.live)
the #!/usr/bin/python at the top enables the script to be called directly, though python must be located there (to confirm, type $which python), and you must set the file to have execute permission ($chmod +x myargs.py)

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