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

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"

其他回答

我的实现下降到子字典,忽略None值,但失败与TypeError如果发现任何其他

def deep_get(d: dict, *keys, default=None):
    """ Safely get a nested value from a dict

    Example:
        config = {'device': None}
        deep_get(config, 'device', 'settings', 'light')
        # -> None
        
    Example:
        config = {'device': True}
        deep_get(config, 'device', 'settings', 'light')
        # -> TypeError

    Example:
        config = {'device': {'settings': {'light': 'bright'}}}
        deep_get(config, 'device', 'settings', 'light')
        # -> 'light'

    Note that it returns `default` is a key is missing or when it's None.
    It will raise a TypeError if a value is anything else but a dict or None.
    
    Args:
        d: The dict to descend into
        keys: A sequence of keys to follow
        default: Custom default value
    """
    # Descend while we can
    try:
        for k in keys:
            d = d[k]
    # If at any step a key is missing, return default
    except KeyError:
        return default
    # If at any step the value is not a dict...
    except TypeError:
        # ... if it's a None, return default. Assume it would be a dict.
        if d is None:
            return default
        # ... if it's something else, raise
        else:
            raise
    # If the value was found, return it
    else:
        return d

通过把所有这些答案和我做的小改变结合起来,我认为这个函数会很有用。安全、快捷、易于维护。

def deep_get(dictionary, keys, default=None):
    return reduce(lambda d, key: d.get(key, default) if isinstance(d, dict) else default, keys.split("."), dictionary)

例子:

from functools import reduce
def deep_get(dictionary, keys, default=None):
    return reduce(lambda d, key: d.get(key, default) if isinstance(d, dict) else default, keys.split("."), dictionary)

person = {'person':{'name':{'first':'John'}}}
print(deep_get(person, "person.name.first"))    # John

print(deep_get(person, "person.name.lastname")) # None

print(deep_get(person, "person.name.lastname", default="No lastname"))  # No lastname

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')

在第一阶段,你可以得到一个空字典。

example_dict.get('key1',{}).get('key2')

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