John O'Hagan wrote:
> I find this surprising:
>
import argparse
parser = argparse.ArgumentParser()
parser.add_argument('--test', action='append', default=['default'])
> [...]
parser.parse_args(['--test', 'arg'])
> Namespace(test=['default', 'arg'])
>
> As an argument is provided, I didn't expect to see the default in there as
> well. From the argparse docs: "the default value is used when the option
> string was not present at the command line". While it doesn't say "_only_
> when...", that's what I would have expected.
>
> Is there a reason for this behaviour? Is there a way to do what I want, to
> get a list of appended options _or_ a default list (apart from setting
> defaults outside the parser, a good way to lose track of things)?
You could either tweak the default value or the action:
$ cat append.py
import argparse
class DefaultList(list):
def __copy__(self):
return []
if __name__ == "__main__":
p = argparse.ArgumentParser()
p.add_argument("-t", "--test", action="append",
default=DefaultList(["default"]))
print p.parse_args()
$ python append.py -t one -t two
Namespace(test=['one', 'two'])
$ python append.py -t one
Namespace(test=['one'])
$ python append.py
Namespace(test=['default'])
$ cat append1.py
import argparse
class DefaultAppend(argparse._AppendAction):
def __call__(self, parser, namespace, values, option_string=None):
items = argparse._copy.copy(argparse._ensure_value(namespace,
self.dest, []))
try:
self._not_first
except AttributeError:
self._not_first = True
del items[:]
items.append(values)
setattr(namespace, self.dest, items)
if __name__ == "__main__":
p = argparse.ArgumentParser()
p.add_argument("-t", "--test", action=DefaultAppend,
default=["default"])
print p.parse_args()
$ python append1.py -t one -t two
Namespace(test=['one', 'two'])
$ python append1.py -t one
Namespace(test=['one'])
$ python append1.py
Namespace(test=['default'])
Peter
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