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

对于嵌套的字典/JSON查找,可以使用dictor

PIP安装指示器

dict对象

{
    "characters": {
        "Lonestar": {
            "id": 55923,
            "role": "renegade",
            "items": [
                "space winnebago",
                "leather jacket"
            ]
        },
        "Barfolomew": {
            "id": 55924,
            "role": "mawg",
            "items": [
                "peanut butter jar",
                "waggy tail"
            ]
        },
        "Dark Helmet": {
            "id": 99999,
            "role": "Good is dumb",
            "items": [
                "Shwartz",
                "helmet"
            ]
        },
        "Skroob": {
            "id": 12345,
            "role": "Spaceballs CEO",
            "items": [
                "luggage"
            ]
        }
    }
}

要获得龙星的物品,只需提供一个点分隔的路径,即

import json
from dictor import dictor

with open('test.json') as data: 
    data = json.load(data)

print dictor(data, 'characters.Lonestar.items')

>> [u'space winnebago', u'leather jacket']

如果键不在路径中,您可以提供回退值

你还有很多选择,比如忽略字母大小写,使用'以外的其他字符。作为路径分隔符,

https://github.com/perfecto25/dictor

其他回答

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)

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)

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

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)

对于二级键检索,你可以这样做:

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