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?
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
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
其他回答
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 ''
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)
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)
如果您想使用另一个库来解决问题,这是最好的方法
https://github.com/maztohir/dict-path
from dict-path import DictPath
data_dict = {
"foo1": "bar1",
"foo2": "bar2",
"foo3": {
"foo4": "bar4",
"foo5": {
"foo6": "bar6",
"foo7": "bar7",
},
}
}
data_dict_path = DictPath(data_dict)
data_dict_path.get('key1/key2/key3')
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)