python - 递归点字典

Question

我有一个实用程序类，它使 Python 字典在获取和设置属性方面的行为有点像 JavaScript 对象。

class DotDict(dict):
    """
    a dictionary that supports dot notation 
    as well as dictionary access notation 
    usage: d = DotDict() or d = DotDict({'val1':'first'})
    set attributes: d.val2 = 'second' or d['val2'] = 'second'
    get attributes: d.val2 or d['val2']
    """
    __getattr__ = dict.__getitem__
    __setattr__ = dict.__setitem__
    __delattr__ = dict.__delitem__

我想这样做，以便它也将嵌套字典转换为 DotDict() 实例。我希望能够用 __init__ 或 __new__ 做这样的事情，但我还没有想出任何可行的方法:

def __init__(self, dct):
    for key in dct.keys():
        if hasattr(dct[key], 'keys'):
            dct[key] = DotDict(dct[key])

如何递归地将嵌套字典转换为 DotDict() 实例？

>>> dct = {'scalar_value':1, 'nested_dict':{'value':2}}
>>> dct = DotDict(dct)

>>> print dct
{'scalar_value': 1, 'nested_dict': {'value': 2}}

>>> print type(dct)
<class '__main__.DotDict'>

>>> print type(dct['nested_dict'])
<type 'dict'>

Answer 1

我没有看到您在构造函数中复制值的位置。因此，这里 DotDict 总是空的。当我添加键分配时，它起作用了:

class DotDict(dict):
    """
    a dictionary that supports dot notation 
    as well as dictionary access notation 
    usage: d = DotDict() or d = DotDict({'val1':'first'})
    set attributes: d.val2 = 'second' or d['val2'] = 'second'
    get attributes: d.val2 or d['val2']
    """
    __getattr__ = dict.__getitem__
    __setattr__ = dict.__setitem__
    __delattr__ = dict.__delitem__

    def __init__(self, dct):
        for key, value in dct.items():
            if hasattr(value, 'keys'):
                value = DotDict(value)
            self[key] = value


dct = {'scalar_value':1, 'nested_dict':{'value':2, 'nested_nested': {'x': 21}}}
dct = DotDict(dct)

print dct.nested_dict.nested_nested.x

它看起来有点危险且容易出错，更不用说给其他开发人员带来无数惊喜的来源了，但它似乎在起作用。