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
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
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 ''
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
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)
因为如果缺少一个键就会引发一个键错误是合理的,我们甚至可以不检查它,让它像这样单一:
def get_dict(d, kl):
cur = d[kl[0]]
return get_dict(cur, kl[1:]) if len(kl) > 1 else cur