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

减少方法的改进很少,使其与列表一起工作。也使用数据路径作为字符串除以点,而不是数组。

def deep_get(dictionary, path):
    keys = path.split('.')
    return reduce(lambda d, key: d[int(key)] if isinstance(d, list) else d.get(key) if d else None, keys, dictionary)

其他回答

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)

我已经编写了一个deepextract包,它完全符合您的要求:https://github.com/ya332/deepextract 你可以这样做

from deepextract import deepextract
# Demo: deepextract.extract_key(obj, key)
deeply_nested_dict = {
    "items": {
        "item": {
            "id": {
                "type": {
                    "donut": {
                        "name": {
                            "batters": {
                                "my_target_key": "my_target_value"
                            }
                        }
                    }
                }
            }
        }
    }
}
print(deepextract.extract_key(deeply_nested_dict, "my_target_key") == "my_target_value")

返回

True

虽然reduce方法简洁而简短,但我认为简单的循环更容易理解。我还包含了一个默认参数。

def deep_get(_dict, keys, default=None):
    for key in keys:
        if isinstance(_dict, dict):
            _dict = _dict.get(key, default)
        else:
            return default
    return _dict

作为理解reduce一行程序如何工作的练习,我执行了以下操作。但最终循环方法对我来说似乎更直观。

def deep_get(_dict, keys, default=None):

    def _reducer(d, key):
        if isinstance(d, dict):
            return d.get(key, default)
        return default

    return reduce(_reducer, keys, _dict)

使用

nested = {'a': {'b': {'c': 42}}}

print deep_get(nested, ['a', 'b'])
print deep_get(nested, ['a', 'b', 'z', 'z'], default='missing')

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

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)