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。
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
对于二级键检索,你可以这样做:
key2_value = (example_dict.get('key1') or {}).get('key2')
对于嵌套的字典/JSON查找,可以使用dictor
PIP安装指示器
dict对象
{
"characters": {
"Lonestar": {
"id": 55923,
"role": "renegade",
"items": [
"space winnebago",
"leather jacket"
]
},
"Barfolomew": {
"id": 55924,
"role": "mawg",
"items": [
"peanut butter jar",
"waggy tail"
]
},
"Dark Helmet": {
"id": 99999,
"role": "Good is dumb",
"items": [
"Shwartz",
"helmet"
]
},
"Skroob": {
"id": 12345,
"role": "Spaceballs CEO",
"items": [
"luggage"
]
}
}
}
要获得龙星的物品,只需提供一个点分隔的路径,即
import json
from dictor import dictor
with open('test.json') as data:
data = json.load(data)
print dictor(data, 'characters.Lonestar.items')
>> [u'space winnebago', u'leather jacket']
如果键不在路径中,您可以提供回退值
你还有很多选择,比如忽略字母大小写,使用'以外的其他字符。作为路径分隔符,
https://github.com/perfecto25/dictor
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)
我建议你试试蟒蛇本尼迪克特。
它是一个dict子类,提供小键盘支持等功能。
安装:pip install python-benedict
from benedict import benedict
example_dict = benedict(example_dict, keypath_separator='.')
现在你可以使用keypath访问嵌套值:
val = example_dict['key1.key2']
# using 'get' method to avoid a possible KeyError:
val = example_dict.get('key1.key2')
或者使用键列表访问嵌套值:
val = example_dict['key1', 'key2']
# using get to avoid a possible KeyError:
val = example_dict.get(['key1', 'key2'])
它在GitHub上经过了很好的测试和开源:
https://github.com/fabiocaccamo/python-benedict
注:我是这个项目的作者