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
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
其他回答
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
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
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。
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
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)