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
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)
还有一个相同功能的函数,也返回一个布尔值来表示是否找到键,并处理一些意想不到的错误。
'''
json : json to extract value from if exists
path : details.detail.first_name
empty path represents root
returns a tuple (boolean, object)
boolean : True if path exists, otherwise False
object : the object if path exists otherwise None
'''
def get_json_value_at_path(json, path=None, default=None):
if not bool(path):
return True, json
if type(json) is not dict :
raise ValueError(f'json={json}, path={path} not supported, json must be a dict')
if type(path) is not str and type(path) is not list:
raise ValueError(f'path format {path} not supported, path can be a list of strings like [x,y,z] or a string like x.y.z')
if type(path) is str:
path = path.strip('.').split('.')
key = path[0]
if key in json.keys():
return get_json_value_at_path(json[key], path[1:], default)
else:
return False, default
使用示例:
my_json = {'details' : {'first_name' : 'holla', 'last_name' : 'holla'}}
print(get_json_value_at_path(my_json, 'details.first_name', ''))
print(get_json_value_at_path(my_json, 'details.phone', ''))
(真的,大声叫) (假的,”)
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
下面是一个基于unutbu函数答案的解决方案:
Python命名指南 默认值作为参数 不用try,只是检查key是否在object上
def safe_get(dictionary, *keys, default=None):
for key in keys:
if key not in dictionary:
return default
dictionary = dictionary[key]
return 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