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
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,”)
其他回答
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)
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
unutbu回答的一个改编,我发现在我自己的代码中很有用:
example_dict.setdefaut('key1', {}).get('key2')
如果key1还没有这个键,它会为它生成一个字典条目,以避免出现KeyError。如果您希望像我这样以包含键对的嵌套字典结束,这似乎是最简单的解决方案。
你也可以使用python reduce:
def deep_get(dictionary, *keys):
return reduce(lambda d, key: d.get(key) if d else None, keys, dictionary)
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)