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
你可以使用pydash:
import pydash as _ #NOTE require `pip install pydash`
_.get(example_dict, 'key1.key2', default='Default')
https://pydash.readthedocs.io/en/latest/api.html
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 ''
def safeget(_dct, *_keys):
if not isinstance(_dct, dict): raise TypeError("Is not instance of dict")
def foo(dct, *keys):
if len(keys) == 0: return dct
elif not isinstance(_dct, dict): return None
else: return foo(dct.get(keys[0], None), *keys[1:])
return foo(_dct, *_keys)
assert safeget(dict()) == dict()
assert safeget(dict(), "test") == None
assert safeget(dict([["a", 1],["b", 2]]),"a", "d") == None
assert safeget(dict([["a", 1],["b", 2]]),"a") == 1
assert safeget({"a":{"b":{"c": 2}},"d":1}, "a", "b")["c"] == 2
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
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)
我使用的一个解决方案类似于double get,但具有使用if else逻辑避免TypeError的额外能力:
value = example_dict['key1']['key2'] if example_dict.get('key1') and example_dict['key1'].get('key2') else default_value
然而,字典嵌套越多,这就变得越麻烦。