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

一个简单的类,可以包装字典,并根据键进行检索:

class FindKey(dict):
    def get(self, path, default=None):
        keys = path.split(".")
        val = None

        for key in keys:
            if val:
                if isinstance(val, list):
                    val = [v.get(key, default) if v else None for v in val]
                else:
                    val = val.get(key, default)
            else:
                val = dict.get(self, key, default)

            if not val:
                break

        return val

例如:

person = {'person':{'name':{'first':'John'}}}
FindDict(person).get('person.name.first') # == 'John'

如果该键不存在,则默认返回None。你可以在FindDict包装器中使用default=键覆盖它,例如':

FindDict(person, default='').get('person.name.last') # == doesn't exist, so ''

其他回答

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')

你可以用虚线:

PIP安装

from dotted.collection import DottedDict

assert DottedDict(dict(foo=dict(bar="baz")))["foo"]["bar"] == "baz"
assert DottedDict(dict(foo=dict(bar="baz")))["foo.bar"] == "baz"
assert DottedDict(dict(foo=dict(bar="baz"))).get("lorem.ipsum", None) is None
assert DottedDict(dict(foo=dict(bar="baz"))).get("lorem.ipsum", "default") == "default"

Glom是一个很好的库,可以进入点查询:

In [1]: from glom import glom

In [2]: data = {'a': {'b': {'c': 'd'}}}

In [3]: glom(data, "a.b.c")
Out[3]: 'd'

查询失败有一个很好的堆栈跟踪,指出确切的故障点:

In [4]: glom(data, "a.b.foo")
---------------------------------------------------------------------------
PathAccessError                           Traceback (most recent call last)
<ipython-input-4-2a3467493ac4> in <module>
----> 1 glom(data, "a.b.foo")

~/.cache/pypoetry/virtualenvs/neural-knapsack-dE7ihQtM-py3.8/lib/python3.8/site-packages/glom/core.py in glom(target, spec, **kwargs)
   2179 
   2180     if err:
-> 2181         raise err
   2182     return ret
   2183 

PathAccessError: error raised while processing, details below.
 Target-spec trace (most recent last):
 - Target: {'a': {'b': {'c': 'd'}}}
 - Spec: 'a.b.foo'
glom.core.PathAccessError: could not access 'foo', part 2 of Path('a', 'b', 'foo'), got error: KeyError('foo')

默认保护:

In [5]: glom(data, "a.b.foo", default="spam")
Out[5]: 'spam'

格洛姆的魅力在于多功能的规格参数。例如,可以很容易地从以下数据中提取所有的名字:

In [8]: data = {
   ...:     "people": [
   ...:         {"first_name": "Alice", "last_name": "Adams"},
   ...:         {"first_name": "Bob", "last_name": "Barker"}
   ...:     ]
   ...: }

In [9]: glom(data, ("people", ["first_name"]))
Out[9]: ['Alice', 'Bob']

更多的例子请阅读glom文档。

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

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)