我想从数据类动态创建Pydantic模型,类似于您可以从数据类动态创建棉花糖模式的方法,就像在marshmallow-dataclass或https://stevenloria.com/dynamic-schemas-in-marshmallow/中一样。有没有一个库或者简单的方法可以做到这一点?
一些背景知识--我更喜欢在我的业务逻辑中使用数据类,而不是直接使用Pydantic模型。我只在我的FastAPI应用程序中使用Pydantic模型来序列化/反序列化带有驼峰式大小写字段的数据。然而,我发现自己基本上重复了数据类的定义,这是很低效的。
示例输入:
from typing import List
from dataclasses import dataclass
@dataclass
class Item:
id: int = None
stuff: str = None
height: float = None
@dataclass
class Bag:
id: int = None
name: str = None
things: List[Item] = None
@dataclass
class Basket:
id: int = None
recipient: str = None
bags: List[Bag] = None
best_item: Item = None所需输出:
from typing import List
from pydantic.main import BaseModel
def camel_case_converter(value: str):
parts = value.lower().split('_')
return parts[0] + ''.join(i.title() for i in parts[1:])
class CamelBaseModel(BaseModel):
class Config:
alias_generator = camel_case_converter
class Item(CamelBaseModel):
id: int = None
stuff: str = None
height: float = None
class Bag(CamelBaseModel):
id: int = None
name: str = None
things: List[Item] = None
class Basket(CamelBaseModel):
id: int = None
recipient: str = None
bags: List[Bag] = None
best_item: Item = None发布于 2021-03-22 20:00:29
也许是这样的?(来自https://github.com/samuelcolvin/pydantic/issues/1967#issuecomment-742698281)
from typing import Type
from pydantic import BaseModel
from pydantic.dataclasses import dataclass as pydantic_dataclass
from typing import List
from dataclasses import dataclass
def model_from_dataclass(kls: 'StdlibDataclass') -> Type[BaseModel]:
"""Converts a stdlib dataclass to a pydantic BaseModel"""
return pydantic_dataclass(kls).__pydantic_model__
@dataclass
class Item:
id: int = None
stuff: str = None
height: float = None
ItemBaseModel = model_from_dataclass(Item)https://stackoverflow.com/questions/65888153
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