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
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
其他回答
下面是一个基于unutbu函数答案的解决方案:
Python命名指南 默认值作为参数 不用try,只是检查key是否在object上
def safe_get(dictionary, *keys, default=None):
for key in keys:
if key not in dictionary:
return default
dictionary = dictionary[key]
return dictionary
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
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)
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)
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 ''