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
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
其他回答
如果您想使用另一个库来解决问题,这是最好的方法
https://github.com/maztohir/dict-path
from dict-path import DictPath
data_dict = {
"foo1": "bar1",
"foo2": "bar2",
"foo3": {
"foo4": "bar4",
"foo5": {
"foo6": "bar6",
"foo7": "bar7",
},
}
}
data_dict_path = DictPath(data_dict)
data_dict_path.get('key1/key2/key3')
unutbu回答的一个改编,我发现在我自己的代码中很有用:
example_dict.setdefaut('key1', {}).get('key2')
如果key1还没有这个键,它会为它生成一个字典条目,以避免出现KeyError。如果您希望像我这样以包含键对的嵌套字典结束,这似乎是最简单的解决方案。
已经有很多很好的答案,但我已经提出了一个类似于JavaScript领域的lodash get的函数,它也支持通过索引进入列表:
def get(value, keys, default_value = None):
'''
Useful for reaching into nested JSON like data
Inspired by JavaScript lodash get and Clojure get-in etc.
'''
if value is None or keys is None:
return None
path = keys.split('.') if isinstance(keys, str) else keys
result = value
def valid_index(key):
return re.match('^([1-9][0-9]*|[0-9])$', key) and int(key) >= 0
def is_dict_like(v):
return hasattr(v, '__getitem__') and hasattr(v, '__contains__')
for key in path:
if isinstance(result, list) and valid_index(key) and int(key) < len(result):
result = result[int(key)] if int(key) < len(result) else None
elif is_dict_like(result) and key in result:
result = result[key]
else:
result = default_value
break
return result
def test_get():
assert get(None, ['foo']) == None
assert get({'foo': 1}, None) == None
assert get(None, None) == None
assert get({'foo': 1}, []) == {'foo': 1}
assert get({'foo': 1}, ['foo']) == 1
assert get({'foo': 1}, ['bar']) == None
assert get({'foo': 1}, ['bar'], 'the default') == 'the default'
assert get({'foo': {'bar': 'hello'}}, ['foo', 'bar']) == 'hello'
assert get({'foo': {'bar': 'hello'}}, 'foo.bar') == 'hello'
assert get({'foo': [{'bar': 'hello'}]}, 'foo.0.bar') == 'hello'
assert get({'foo': [{'bar': 'hello'}]}, 'foo.1') == None
assert get({'foo': [{'bar': 'hello'}]}, 'foo.1.bar') == None
assert get(['foo', 'bar'], '1') == 'bar'
assert get(['foo', 'bar'], '2') == None
对于二级键检索,你可以这样做:
key2_value = (example_dict.get('key1') or {}).get('key2')
我改编了GenesRus和unutbu的答案,非常简单:
class new_dict(dict):
def deep_get(self, *args, default=None):
_empty_dict = {}
out = self
for key in args:
out = out.get(key, _empty_dict)
return out if out else default
它适用于:
d = new_dict(some_data)
d.deep_get("key1", "key2", "key3", ..., default=some_value)