unutbu回答的一个改编，我发现在我自己的代码中很有用:

example_dict.setdefaut('key1', {}).get('key2')

如果key1还没有这个键，它会为它生成一个字典条目，以避免出现KeyError。如果您希望像我这样以包含键对的嵌套字典结束，这似乎是最简单的解决方案。

我的实现下降到子字典，忽略None值，但失败与TypeError如果发现任何其他

def deep_get(d: dict, *keys, default=None):
    """ Safely get a nested value from a dict

    Example:
        config = {'device': None}
        deep_get(config, 'device', 'settings', 'light')
        # -> None
        
    Example:
        config = {'device': True}
        deep_get(config, 'device', 'settings', 'light')
        # -> TypeError

    Example:
        config = {'device': {'settings': {'light': 'bright'}}}
        deep_get(config, 'device', 'settings', 'light')
        # -> 'light'

    Note that it returns `default` is a key is missing or when it's None.
    It will raise a TypeError if a value is anything else but a dict or None.
    
    Args:
        d: The dict to descend into
        keys: A sequence of keys to follow
        default: Custom default value
    """
    # Descend while we can
    try:
        for k in keys:
            d = d[k]
    # If at any step a key is missing, return default
    except KeyError:
        return default
    # If at any step the value is not a dict...
    except TypeError:
        # ... if it's a None, return default. Assume it would be a dict.
        if d is None:
            return default
        # ... if it's something else, raise
        else:
            raise
    # If the value was found, return it
    else:
        return d

从Python 3.4开始，你可以使用suppress (KeyError)来访问嵌套的json对象，而不用担心KeyError

from contextlib import suppress

with suppress(KeyError):
    a1 = json_obj['key1']['key2']['key3']
    a2 = json_obj['key4']['key5']['key6']
    a3 = json_obj['key7']['key8']['key9']

Techdragon提供。看看他的回答，了解更多细节:https://stackoverflow.com/a/45874251/1189659

一个简单的类，可以包装字典，并根据键进行检索:

class FindKey(dict):
    def get(self, path, default=None):
        keys = path.split(".")
        val = None

        for key in keys:
            if val:
                if isinstance(val, list):
                    val = [v.get(key, default) if v else None for v in val]
                else:
                    val = val.get(key, default)
            else:
                val = dict.get(self, key, default)

            if not val:
                break

        return val

例如:

person = {'person':{'name':{'first':'John'}}}
FindDict(person).get('person.name.first') # == 'John'

如果该键不存在，则默认返回None。你可以在FindDict包装器中使用default=键覆盖它，例如':

FindDict(person, default='').get('person.name.last') # == doesn't exist, so ''

对于二级键检索，你可以这样做:

key2_value = (example_dict.get('key1') or {}).get('key2')

我建议你试试蟒蛇本尼迪克特。

它是一个dict子类，提供小键盘支持等功能。

安装:pip install python-benedict

from benedict import benedict

example_dict = benedict(example_dict, keypath_separator='.')

现在你可以使用keypath访问嵌套值:

val = example_dict['key1.key2']

# using 'get' method to avoid a possible KeyError:
val = example_dict.get('key1.key2')

或者使用键列表访问嵌套值:

val = example_dict['key1', 'key2']

# using get to avoid a possible KeyError:
val = example_dict.get(['key1', 'key2'])

它在GitHub上经过了很好的测试和开源:

https://github.com/fabiocaccamo/python-benedict

注:我是这个项目的作者