获得点访问(但不是数组访问)的一个简单方法是在Python中使用一个普通对象。是这样的:

class YourObject:
    def __init__(self, *args, **kwargs):
        for k, v in kwargs.items():
            setattr(self, k, v)

.．.像这样使用它:

>>> obj = YourObject(key="value")
>>> print(obj.key)
"value"

．.． 把它转换成字典:

>>> print(obj.__dict__)
{"key": "value"}

如果你已经在使用pandas，你可以构造一个pandas Series或DataFrame，从中你可以通过点语法访问项目:

1级字典:

import pandas as pd

my_dictionary = pd.Series({
  'key1': 'value1',
  'key2': 'value2'
})

print(my_dictionary.key1)
# Output: value1

2级字典:

import pandas as pd

my_dictionary = pd.DataFrame({
  'key1': {
    'inner_key1': 'value1'
  },
  'key2': {
    'inner_key2': 'value2'
  }
})

print(my_dictionary.key1.inner_key1)
# Output: value1

请注意，这可能在规范化数据结构(其中每个字典条目都具有相同的结构)下工作得更好。在上面的第二个例子中，得到的DataFrame是:

              key1    key2
inner_key1  value1     NaN
inner_key2     NaN  value2

我不喜欢在(超过)10年前的火灾中添加另一个日志，但我也会检查dotwiz库，它是我最近发布的——实际上就在今年。

它是一个相对较小的库，在基准测试中，它在get(访问)和设置(创建)时间方面也表现得非常好，至少与其他备选方案相比是这样。

通过pip安装dotwiz

pip install dotwiz

它能做你想让它做的所有事情，并继承dict的子类，所以它的操作就像一个普通的字典:

from dotwiz import DotWiz

dw = DotWiz()
dw.hello = 'world'
dw.hello
dw.hello += '!'
# dw.hello and dw['hello'] now both return 'world!'
dw.val = 5
dw.val2 = 'Sam'

最重要的是，你可以将它转换为dict对象:

d = dw.to_dict()
dw = DotWiz(d) # automatic conversion in constructor

这意味着如果你想访问的东西已经是dict形式的，你可以把它变成一个dotwz来方便访问:

import json
json_dict = json.loads(text)
data = DotWiz(json_dict)
print data.location.city

最后，我正在做的一些令人兴奋的事情是一个现有的特性请求，这样它就会自动创建新的子DotWiz实例，这样你就可以做这样的事情:

dw = DotWiz()
dw['people.steve.age'] = 31

dw
# ✫(people=✫(steve=✫(age=31)))

与点图比较

我在下面添加了一个快速而粗略的性能比较。

首先，用pip安装两个库:

pip install dotwiz dotmap

为了进行基准测试，我编写了以下代码:

from timeit import timeit

from dotwiz import DotWiz
from dotmap import DotMap


d = {'hey': {'so': [{'this': {'is': {'pretty': {'cool': True}}}}]}}

dw = DotWiz(d)
# ✫(hey=✫(so=[✫(this=✫(is=✫(pretty={'cool'})))]))

dm = DotMap(d)
# DotMap(hey=DotMap(so=[DotMap(this=DotMap(is=DotMap(pretty={'cool'})))]))

assert dw.hey.so[0].this['is'].pretty.cool == dm.hey.so[0].this['is'].pretty.cool

n = 100_000

print('dotwiz (create):  ', round(timeit('DotWiz(d)', number=n, globals=globals()), 3))
print('dotmap (create):  ', round(timeit('DotMap(d)', number=n, globals=globals()), 3))
print('dotwiz (get):  ', round(timeit("dw.hey.so[0].this['is'].pretty.cool", number=n, globals=globals()), 3))
print('dotmap (get):  ', round(timeit("dm.hey.so[0].this['is'].pretty.cool", number=n, globals=globals()), 3))

结果，在我的M1 Mac上运行Python 3.10:

dotwiz (create):   0.189
dotmap (create):   1.085
dotwiz (get):   0.014
dotmap (get):   0.335

基于Kugel的回答，并考虑到Mike Graham的警告，如果我们制作一个包装器呢?

class DictWrap(object):
  """ Wrap an existing dict, or create a new one, and access with either dot 
    notation or key lookup.

    The attribute _data is reserved and stores the underlying dictionary.
    When using the += operator with create=True, the empty nested dict is 
    replaced with the operand, effectively creating a default dictionary
    of mixed types.

    args:
      d({}): Existing dict to wrap, an empty dict is created by default
      create(True): Create an empty, nested dict instead of raising a KeyError

    example:
      >>>dw = DictWrap({'pp':3})
      >>>dw.a.b += 2
      >>>dw.a.b += 2
      >>>dw.a['c'] += 'Hello'
      >>>dw.a['c'] += ' World'
      >>>dw.a.d
      >>>print dw._data
      {'a': {'c': 'Hello World', 'b': 4, 'd': {}}, 'pp': 3}

  """

  def __init__(self, d=None, create=True):
    if d is None:
      d = {}
    supr = super(DictWrap, self)  
    supr.__setattr__('_data', d)
    supr.__setattr__('__create', create)

  def __getattr__(self, name):
    try:
      value = self._data[name]
    except KeyError:
      if not super(DictWrap, self).__getattribute__('__create'):
        raise
      value = {}
      self._data[name] = value

    if hasattr(value, 'items'):
      create = super(DictWrap, self).__getattribute__('__create')
      return DictWrap(value, create)
    return value

  def __setattr__(self, name, value):
    self._data[name] = value  

  def __getitem__(self, key):
    try:
      value = self._data[key]
    except KeyError:
      if not super(DictWrap, self).__getattribute__('__create'):
        raise
      value = {}
      self._data[key] = value

    if hasattr(value, 'items'):
      create = super(DictWrap, self).__getattribute__('__create')
      return DictWrap(value, create)
    return value

  def __setitem__(self, key, value):
    self._data[key] = value

  def __iadd__(self, other):
    if self._data:
      raise TypeError("A Nested dict will only be replaced if it's empty")
    else:
      return other

kaggle_environments使用的实现是一个名为structify的函数。

class Struct(dict):
    def __init__(self, **entries):
        entries = {k: v for k, v in entries.items() if k != "items"}
        dict.__init__(self, entries)
        self.__dict__.update(entries)

    def __setattr__(self, attr, value):
        self.__dict__[attr] = value
        self[attr] = value

# Added benefit of cloning lists and dicts.
def structify(o):
    if isinstance(o, list):
        return [structify(o[i]) for i in range(len(o))]
    elif isinstance(o, dict):
        return Struct(**{k: structify(v) for k, v in o.items()})
    return o

https://github.com/Kaggle/kaggle-environments/blob/master/kaggle_environments/utils.py

这可能有助于在《ConnectX》等游戏中测试AI模拟代理

from kaggle_environments import structify

obs  = structify({ 'remainingOverageTime': 60, 'step': 0, 'mark': 1, 'board': [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]})
conf = structify({ 'timeout': 2, 'actTimeout': 2, 'agentTimeout': 60, 'episodeSteps': 1000, 'runTimeout': 1200, 'columns': 7, 'rows': 6, 'inarow': 4, '__raw_path__': '/kaggle_simulations/agent/main.py' })

def agent(obs, conf):
  action = obs.step % conf.columns
  return action

我最近遇到了“Box”库，它也做同样的事情。

安装命令:pip install python-box

例子:

from box import Box

mydict = {"key1":{"v1":0.375,
                    "v2":0.625},
          "key2":0.125,
          }
mydict = Box(mydict)

print(mydict.key1.v1)

我发现它比其他现有的库(如dotmap)更有效，当你有大量嵌套字典时，dotmap会产生python递归错误。

链接到图书馆和详细信息:https://pypi.org/project/python-box/