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

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)

其他回答

在深入获取属性后,我使用点表示法安全地获得嵌套的dict值。这适用于我,因为我的字典是反序列化的MongoDB对象,所以我知道键名不包含.s。此外,在我的上下文中,我可以指定一个数据中没有的虚假回退值(None),因此在调用函数时可以避免使用try/except模式。

from functools import reduce # Python 3
def deepgetitem(obj, item, fallback=None):
    """Steps through an item chain to get the ultimate value.

    If ultimate value or path to value does not exist, does not raise
    an exception and instead returns `fallback`.

    >>> d = {'snl_final': {'about': {'_icsd': {'icsd_id': 1}}}}
    >>> deepgetitem(d, 'snl_final.about._icsd.icsd_id')
    1
    >>> deepgetitem(d, 'snl_final.about._sandbox.sbx_id')
    >>>
    """
    def getitem(obj, name):
        try:
            return obj[name]
        except (KeyError, TypeError):
            return fallback
    return reduce(getitem, item.split('.'), obj)

还有一个相同功能的函数,也返回一个布尔值来表示是否找到键,并处理一些意想不到的错误。

'''
json : json to extract value from if exists
path : details.detail.first_name
            empty path represents root

returns a tuple (boolean, object)
        boolean : True if path exists, otherwise False
        object : the object if path exists otherwise None

'''
def get_json_value_at_path(json, path=None, default=None):

    if not bool(path):
        return True, json
    if type(json) is not dict :
        raise ValueError(f'json={json}, path={path} not supported, json must be a dict')
    if type(path) is not str and type(path) is not list:
        raise ValueError(f'path format {path} not supported, path can be a list of strings like [x,y,z] or a string like x.y.z')

    if type(path) is str:
        path = path.strip('.').split('.')
    key = path[0]
    if key in json.keys():
        return get_json_value_at_path(json[key], path[1:], default)
    else:
        return False, default

使用示例:

my_json = {'details' : {'first_name' : 'holla', 'last_name' : 'holla'}}
print(get_json_value_at_path(my_json, 'details.first_name', ''))
print(get_json_value_at_path(my_json, 'details.phone', ''))

(真的,大声叫) (假的,”)

我使用的一个解决方案类似于double get,但具有使用if else逻辑避免TypeError的额外能力:

    value = example_dict['key1']['key2'] if example_dict.get('key1') and example_dict['key1'].get('key2') else default_value

然而,字典嵌套越多,这就变得越麻烦。

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

你可以使用pydash:

import pydash as _  #NOTE require `pip install pydash`

_.get(example_dict, 'key1.key2', default='Default')

https://pydash.readthedocs.io/en/latest/api.html