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

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

其他回答

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)

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

对于二级键检索,你可以这样做:

key2_value = (example_dict.get('key1') or {}).get('key2')

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。