I want to open file for reading using argparse. In cmd it must look like: my_program.py /filepath
That's my try:
parser = argparse.ArgumentParser()
parser.add_argument('file', type = file)
args = parser.parse_args()
This gives me [edit: added later from comment by OP]:
parser.add_argument('file', type = file) NameError: name 'file' is not defined
Note: argparse.FileType, which this answer mentions, has now been deprecated. See gh-58032: argparse.FileType opens a file and never closes it.
Take a look at the documentation: https://docs.python.org/3/library/argparse.html#type
import argparse
parser = argparse.ArgumentParser()
parser.add_argument('file', type=argparse.FileType('r'))
args = parser.parse_args()
print(args.file.readlines())
argparse.FileType. For example, it leaves an open file handle dangling which is in my opinion sloppy implementation. Usually I would rather be handling the file open() myself so that I can do it properly (with a context manager), and also customise the behaviour and output for 'file not found' errors.
with args.file as f: ... which will close the file earlier in the good scenario case.
argparse automatically sets file to <stdin> (or <stdout>, depending on mode) if the argument is missing or is -. Which is a very cool feature, but if with args.file as f: ... is used, it will (or try to) close them, when it shouldn't.The type of the argument should be string (which is default anyway). So make it like this:
parser = argparse.ArgumentParser()
parser.add_argument('filename')
args = parser.parse_args()
with open(args.filename) as file:
# do stuff here
file since that is a built-in.file is never used in python these days - it was removed in python3, and in python2 the docstring even advises to use open instead. So it's OK to stomp this name, it is even done several times in core libraries.In order to have the file closed gracefully, you can combine argparse.FileType with the with statement
parser.add_argument('file', type=argparse.FileType('r'))
args = parser.parse_args()
with args.file as file:
print(file.read())
with args.file the "file" refers to the name of the argument right? If I had named the argument as -f then it should be with args.-f right?? Correct me if I'm wrong. Argparse beginner hereadd_argument('-f', ..., then you need with args.f as file without the dash
type=argparse.FileType('r'), the file is opened from the point you call parser.parse_args(). Does this mean the file would still be left open if your code crashes before your context handler? Not a big deal, but something to keep in mind.argparse.FileType has been deprecated in Python 3.14.I'll just add the option to use pathlib:
import argparse, pathlib
parser = argparse.ArgumentParser()
parser.add_argument('file', type=pathlib.Path)
args = parser.parse_args()
with args.file.open('r') as file:
print(file.read())
parser.add_argument(
'-f', '--file',
help='JSON input file',
type=argparse.FileType('r'),
)
Simple command line script to reformat JSON files
reformat-json \
-f package.json \
--indent=2 \
--sort-keys \
--output=sorted_package.json
can be code in Python as follows
#!/usr/bin/env python3
import argparse, json, sys
EXIT_SUCCESS = 0
EXIT_FAILURE = 1
def main():
parser = argparse.ArgumentParser()
parser.add_argument(
'-f', '--file',
help='JSON input file',
type=argparse.FileType('r'),
)
parser.add_argument(
'-i', '--indent',
help='Non-negative integer indent level',
type=int
)
parser.add_argument(
'-o', '--output',
help='Write JSON into output file',
type=argparse.FileType('w'),
)
parser.add_argument(
'-s', '--sort-keys',
action='store_true',
help='Sort output JSON by keys',
)
args = parser.parse_args()
if not args.file:
parser.print_usage()
return sys.exit(EXIT_FAILURE)
gson = json.dumps(
json.load(args.file),
indent=args.indent,
sort_keys=args.sort_keys
)
args.file.close()
if args.output:
args.output.write(gson)
args.output.write('\n')
args.output.close()
else:
print(gson)
return sys.exit(EXIT_SUCCESS)
if __name__ == '__main__':
main()
argparse.FileType has been deprecated in Python 3.14.The following avoids the problems with closing files correctly that are present in ilent2's answer. It works with multiple file arguments and resembles the argparse.FileType approach.
import argparse
import contextlib
class FileStack(contextlib.ExitStack):
def file(self, mode):
def _open(path):
obj = open(path, mode)
self.enter_context(obj)
return obj
return _open
with FileStack() as stack:
parser = argparse.ArgumentParser()
parser.add_argument("file", type=stack.file("r"))
args = parser.parse_args()
args.file.read()
# Here, args.file is correctly closed
This implementation allows the "file name" parameter to be optional, as well as giving a short description if and when the user enters the -h or --help argument.
parser = argparse.ArgumentParser(description='Foo is a program that does things')
parser.add_argument('filename', nargs='?')
args = parser.parse_args()
if args.filename is not None:
print('The file name is {}'.format(args.filename))
else:
print('Oh well ; No args, no problems')
type = fncworks iffncis a function that takes a string, and returns a desired object, or raises an error. There isn't afilefunction in Python.type=opendoes work since there is a Python functionopen(filename)'. argparse provides a better function,argparse.FileType()` (actually a function constructor). But a simple string argument as inwims answer is a better starting place.type=filedoes work in Python2.7 (and earlier). But in Python3,filehas been removed. stackoverflow.com/questions/112970/…