将嵌套的Python字典转换为对象? [英] Convert nested Python dict to object?

问题描述

我正在寻找一种优雅的方式来通过对具有一些嵌套的字典和列表(即javascript样式的对象语法)的字典进行属性访问来获取数据.

I'm searching for an elegant way to get data using attribute access on a dict with some nested dicts and lists (i.e. javascript-style object syntax).

例如:

>>> d = {'a': 1, 'b': {'c': 2}, 'd': ["hi", {'foo': "bar"}]}

应该以这种方式访问​​:

Should be accessible in this way:

>>> x = dict2obj(d)
>>> x.a
1
>>> x.b.c
2
>>> x.d[1].foo
bar

我认为，没有递归是不可能的，但是获得字典对象样式的一种好方法是什么?

I think, this is not possible without recursion, but what would be a nice way to get an object style for dicts?

推荐答案

更新:在Python 2.6及更高版本中，请考虑

Update: In Python 2.6 and onwards, consider whether the namedtuple data structure suits your needs:

>>> from collections import namedtuple
>>> MyStruct = namedtuple('MyStruct', 'a b d')
>>> s = MyStruct(a=1, b={'c': 2}, d=['hi'])
>>> s
MyStruct(a=1, b={'c': 2}, d=['hi'])
>>> s.a
1
>>> s.b
{'c': 2}
>>> s.c
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'MyStruct' object has no attribute 'c'
>>> s.d
['hi']

替代方法(原始答案内容)是:

The alternative (original answer contents) is:

class Struct:
    def __init__(self, **entries):
        self.__dict__.update(entries)

然后，您可以使用:

>>> args = {'a': 1, 'b': 2}
>>> s = Struct(**args)
>>> s
<__main__.Struct instance at 0x01D6A738>
>>> s.a
1
>>> s.b
2

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