Python 使用无效的语法名称创建PYDANIC计算字段

这看起来很像是您更愿意应用区别对待的联合模式.请参阅以下示例:

from pydantic import BaseModel, Field
from typing import Literal, Annotated


example = {
    "$type": "Menu",
    "name": "lunch",
    "children": [
        {"$type": "Pasta", "title": "carbonara"},
        {"$type": "Meat", "is_vegetable": False},
    ]
}


class Pasta(BaseModel):
    type: Literal["Pasta"] = Field("Pasta", alias="$type")
    title: str

class Meat(BaseModel):
    type: Literal["Meat"] = Field("Meat", alias="$type")
    is_vegetable: bool


AnyDish = Annotated[Pasta | Meat, Field(discriminator="type")]

class Menu(BaseModel):
    name: str
    children: list[AnyDish]


menu = Menu.model_validate(example)
print(menu)

menu.model_dump(by_alias=True)

它打印:

name='lunch' children=[Pasta(type='Pasta', title='carbonara'), Meat(type='Meat', is_vegetable=False)]
{'name': 'lunch', 'children': [{'$type': 'Pasta', 'title': 'carbonara'}, {'$type': 'Meat', 'is_vegetable': False}]}

这种模式有多个优点:

• 它将类名从标记中分离出来.这通常是首选的，因为类名可能会更改.但你可能仍然想要阅读旧文件.
• 该模式是可扩展和显式的.以后你可以很容易地添加新的菜肴类型，只需将它们包括在AnyDish种菜肴中即可.

为了减少冗长，人们还可以引入一条捷径，比如:

class TypeLiteral:
    def __class_getitem__(cls, tag: str):
        return Annotated[Literal[tag], Field(default=tag, alias="$type")]


class Pasta(BaseModel):
    type: TypeLiteral["Pasta"]
    title: str

class Meat(BaseModel):
    type: TypeLiteral["Meat"]
    is_vegetable: bool

你可以在PYDING DOCKS中找到更多关于歧视unions 的信息:https://docs.pydantic.dev/latest/concepts/unions/#discriminated-unions

我希望这是有用的！