在第一阶段，你可以得到一个空字典。

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

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)

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。

我稍微改变了一下答案。我添加了检查，如果我们使用列表与数字。 所以现在我们可以用任何一种方法。deep_get(allTemp，[0]，{})或deep_get(getMinimalTemp， [0, minimalTemperatureKey]， 26)等

def deep_get(_dict, keys, default=None):
    def _reducer(d, key):
        if isinstance(d, dict):
            return d.get(key, default)
        if isinstance(d, list):
            return d[key] if len(d) > 0 else default
        return default
    return reduce(_reducer, keys, _dict)

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

因为如果缺少一个键就会引发一个键错误是合理的，我们甚至可以不检查它，让它像这样单一:

def get_dict(d, kl):
  cur = d[kl[0]]
  return get_dict(cur, kl[1:]) if len(kl) > 1 else cur