我使用命名元组编写了一个实现,因为它有点易读,所以我认为这很不错。它也可以处理字典嵌套的情况。解析器代码如下:
from collections import namedtuple
class Dict2ObjParser:
def __init__(self, nested_dict):
self.nested_dict = nested_dict
def parse(self):
nested_dict = self.nested_dict
if (obj_type := type(nested_dict)) is not dict:
raise TypeError(f"Expected 'dict' but found '{obj_type}'")
return self._transform_to_named_tuples("root", nested_dict)
def _transform_to_named_tuples(self, tuple_name, possibly_nested_obj):
if type(possibly_nested_obj) is dict:
named_tuple_def = namedtuple(tuple_name, possibly_nested_obj.keys())
transformed_value = named_tuple_def(
*[
self._transform_to_named_tuples(key, value)
for key, value in possibly_nested_obj.items()
]
)
elif type(possibly_nested_obj) is list:
transformed_value = [
self._transform_to_named_tuples(f"{tuple_name}_{i}", possibly_nested_obj[i])
for i in range(len(possibly_nested_obj))
]
else:
transformed_value = possibly_nested_obj
return transformed_value
我使用以下代码测试了基本情况:
x = Dict2ObjParser({
"a": {
"b": 123,
"c": "Hello, World!"
},
"d": [
1,
2,
3
],
"e": [
{
"f": "",
"g": None
},
{
"f": "Foo",
"g": "Bar"
},
{
"h": "Hi!",
"i": None
}
],
"j": 456,
"k": None
}).parse()
print(x)
它会输出以下内容:
root(a=a(b=123, c='Hello, World!'), d=[1, 2, 3], e=[e_0(f='', g=None), e_1(f='Foo', g='Bar'), e_2(h='Hi!', i=None)], j=456, k=None)
稍微格式化后,它看起来像这样:
root(
a=a(
b=123,
c='Hello, World!'
),
d=[1, 2, 3],
e=[
e_0(
f='',
g=None
),
e_1(
f='Foo',
g='Bar'
),
e_2(
h='Hi!',
i=None
)
],
j=456,
k=None
)
我可以像访问其他对象一样访问嵌套字段:
print(x.a.b)
在您的情况下,代码最终将如下所示:
import yaml
with open(file_path, "r") as stream:
nested_dict = yaml.safe_load(stream)
nested_objt = Dict2ObjParser(nested_dict).parse()
我希望这能帮到您!
python-box
。 - evolved