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
我稍微改变了一下答案。我添加了检查,如果我们使用列表与数字。 所以现在我们可以用任何一种方法。deep_get(allTemp,[0],{})或deep_get(getMinimalTemp, [0, minimalTemperatureKey], 26)等
def deep_get(_dict, keys, default=None):
def _reducer(d, key):
if isinstance(d, dict):
return d.get(key, default)
if isinstance(d, list):
return d[key] if len(d) > 0 else default
return default
return reduce(_reducer, keys, _dict)
其他回答
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
从Python 3.4开始,你可以使用suppress (KeyError)来访问嵌套的json对象,而不用担心KeyError
from contextlib import suppress
with suppress(KeyError):
a1 = json_obj['key1']['key2']['key3']
a2 = json_obj['key4']['key5']['key6']
a3 = json_obj['key7']['key8']['key9']
Techdragon提供。看看他的回答,了解更多细节:https://stackoverflow.com/a/45874251/1189659
我已经编写了一个deepextract包,它完全符合您的要求:https://github.com/ya332/deepextract 你可以这样做
from deepextract import deepextract
# Demo: deepextract.extract_key(obj, key)
deeply_nested_dict = {
"items": {
"item": {
"id": {
"type": {
"donut": {
"name": {
"batters": {
"my_target_key": "my_target_value"
}
}
}
}
}
}
}
}
print(deepextract.extract_key(deeply_nested_dict, "my_target_key") == "my_target_value")
返回
True
减少方法的改进很少,使其与列表一起工作。也使用数据路径作为字符串除以点,而不是数组。
def deep_get(dictionary, path):
keys = path.split('.')
return reduce(lambda d, key: d[int(key)] if isinstance(d, list) else d.get(key) if d else None, keys, dictionary)
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)