将argparse.Namespace()视为字典的正确方法是什么?

问题描述

如果我想使用 argparse.ArgumentParser()的结果(这是一个 Namespace 对象)，并使用需要字典或类似映射的对象的方法(参见 collections.Mapping )，什么是正确的方法它吗?

If I want to use the results of argparse.ArgumentParser(), which is a Namespace object, with a method that expects a dictionary or mapping-like object (see collections.Mapping), what is the right way to do it?

C:\>python
Python 2.7.3 (default, Apr 10 2012, 23:31:26) [MSC v.1500 32 bit (Intel)] on win
32
Type "help", "copyright", "credits" or "license" for more information.
>>> import argparse
>>> args = argparse.Namespace()
>>> args.foo = 1
>>> args.bar = [1,2,3]
>>> args.baz = 'yippee'
>>> args['baz']
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'Namespace' object has no attribute '__getitem__'
>>> dir(args)
['__class__', '__contains__', '__delattr__', '__dict__', '__doc__', '__eq__', '_
_format__', '__getattribute__', '__hash__', '__init__', '__module__', '__ne__',
'__new__', '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__sizeof__
', '__str__', '__subclasshook__', '__weakref__', '_get_args', '_get_kwargs', 'ba
r', 'baz', 'foo']

进入"对象并使用其 __ dict __ 属性是否合适?

Is it proper to "reach into" an object and use its __dict__ property?

我认为答案是否定的: __ dict __ 闻起来像实现的约定，而不是接口的含义，就像 __ getattribute __ 或 __ setattr __ 或 __ contains __ 似乎是

I would think the answer is no: __dict__ smells like a convention for implementation, but not for an interface, the way __getattribute__ or __setattr__ or __contains__ seem to be.

推荐答案

您可以使用 vars() :

You can access the namespace's dictionary with vars():

>>> import argparse
>>> args = argparse.Namespace()
>>> args.foo = 1
>>> args.bar = [1,2,3]
>>> d = vars(args)
>>> d
{'foo': 1, 'bar': [1, 2, 3]}

您可以根据需要直接修改字典:

You can modify the dictionary directly if you wish:

>>> d['baz'] = 'store me'
>>> args.baz
'store me'

是的，可以访问__dict__属性.这是定义明确，经过测试且有保证的行为.

Yes, it is okay to access the __dict__ attribute. It is a well-defined, tested, and guaranteed behavior.

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