递归解。它不是最有效的，但我发现它比其他示例更具可读性，而且它不依赖于functools。

def deep_get(d, keys):
    if not keys or d is None:
        return d
    return deep_get(d.get(keys[0]), keys[1:])

例子

d = {'meta': {'status': 'OK', 'status_code': 200}}
deep_get(d, ['meta', 'status_code'])     # => 200
deep_get(d, ['garbage', 'status_code'])  # => None

一个更精致的版本

def deep_get(d, keys, default=None):
    """
    Example:
        d = {'meta': {'status': 'OK', 'status_code': 200}}
        deep_get(d, ['meta', 'status_code'])          # => 200
        deep_get(d, ['garbage', 'status_code'])       # => None
        deep_get(d, ['meta', 'garbage'], default='-') # => '-'
    """
    assert type(keys) is list
    if d is None:
        return default
    if not keys:
        return d
    return deep_get(d.get(keys[0]), keys[1:], default)

如果您想使用另一个库来解决问题，这是最好的方法

https://github.com/maztohir/dict-path

from dict-path import DictPath

data_dict = {
  "foo1": "bar1",
  "foo2": "bar2",
  "foo3": {
     "foo4": "bar4",
     "foo5": {
        "foo6": "bar6",
        "foo7": "bar7",
     },
  }
}

data_dict_path = DictPath(data_dict)
data_dict_path.get('key1/key2/key3')

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

根据Yoav的回答，一个更安全的方法是:

def deep_get(dictionary, *keys):
    return reduce(lambda d, key: d.get(key, None) if isinstance(d, dict) else None, keys, dictionary)

下面是一个基于unutbu函数答案的解决方案:

Python命名指南 默认值作为参数 不用try，只是检查key是否在object上

def safe_get(dictionary, *keys, default=None):
    for key in keys:
        if key not in dictionary:
            return default
        dictionary = dictionary[key]
    return dictionary

我改编了GenesRus和unutbu的答案，非常简单:

class new_dict(dict):
    def deep_get(self, *args, default=None):
        _empty_dict = {}
        out = self
        for key in args:
            out = out.get(key, _empty_dict)
        return out if out else default

它适用于:

d = new_dict(some_data)
d.deep_get("key1", "key2", "key3", ..., default=some_value)