我有一本嵌套的字典。是否只有一种方法可以安全地传递价值观?

try:
    example_dict['key1']['key2']
except KeyError:
    pass

或者python有一个类似get()的方法用于嵌套字典?


当前回答

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

其他回答

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')

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)

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

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)

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