Pydantic -如何对内置类型进行子类划分

Question

我正在尝试创建一个timedelta的子类，它期望接收毫秒而不是秒，但它目前不起作用。

我是在反对吗？有没有一种“正确”的方法来实现这一点与皮丹蒂克？还是我需要告诉皮丹蒂克，MillisecondTimedelta只是一个timedelta。

from datetime import timedelta

from pydantic import BaseModel


class MillisecondTimedelta(timedelta):
    @classmethod
    def __get_validators__(cls):
        # timedelta expects seconds
        yield lambda v: v / 1000
        yield cls


class MyModel(BaseModel):
    td: MillisecondTimedelta

data = {
    "td": 7598040,
}

print(MyModel(**data))

在以下方面的成果：

Traceback (most recent call last):
  File "main.py", line 14, in <module>
    class MyModel(BaseModel):
  File "pydantic/main.py", line 262, in pydantic.main.ModelMetaclass.__new__
  File "pydantic/fields.py", line 315, in pydantic.fields.ModelField.infer
  File "pydantic/fields.py", line 284, in pydantic.fields.ModelField.__init__
  File "pydantic/fields.py", line 362, in pydantic.fields.ModelField.prepare
  File "pydantic/fields.py", line 541, in pydantic.fields.ModelField.populate_validators
  File "pydantic/class_validators.py", line 255, in pydantic.class_validators.prep_validators
  File "pydantic/class_validators.py", line 238, in pydantic.class_validators.make_generic_validator
  File "/usr/lib/python3.8/inspect.py", line 3105, in signature
    return Signature.from_callable(obj, follow_wrapped=follow_wrapped)
  File "/usr/lib/python3.8/inspect.py", line 2854, in from_callable
    return _signature_from_callable(obj, sigcls=cls,
  File "/usr/lib/python3.8/inspect.py", line 2384, in _signature_from_callable
    raise ValueError(
ValueError: no signature found for builtin type <class '__main__.MillisecondTimedelta'>

Answer 1

啊哈，原来我需要改变两件事。

1. 实际上，从验证器返回一个timedelta，我返回的是cls，这是我的自定义子类。
2. 不要子类timedelta

from datetime import timedelta

from pydantic import BaseModel


class MillisecondTimedelta:

    @classmethod
    def __get_validators__(cls):
        yield lambda v: timedelta(milliseconds=v)


class MyModel(BaseModel):
    td: MillisecondTimedelta


data = {
    "td": 7598040,
}

print(repr(MyModel(**data)))

MyModel(td=datetime.timedelta(seconds=7598, microseconds=40000))

Answer 2

如__get_validators__()所示，您需要生成一个或多个验证器。

下面报告了修改后的类；问题是Pydantic理解(对于timedelta字段) int，并以秒(来源)形式浮动。

class MillisecondTimedelta(timedelta):
    @classmethod
    def __get_validators__(cls):
        yield cls.validate

    @classmethod
    def validate(cls, v):
        if any(isinstance(v, t) for t in (int, float)):
            return cls(milliseconds=v)

现在一切都应该正常工作。

>>> data = {"td": 1000}
>>> print(MyModel(**data))
td=MillisecondTimedelta(seconds=1)

编辑:如果没有自定义类和验证器，就可以使用一个函数来编辑分配给类构造函数的值；它需要修饰这个函数，如图所示

class MyModel(BaseModel):
    td: timedelta

    @validator('td')
    def convert_to_ms(cls, v):
        return v / 1000

此外，这一解决方案也在发挥作用：

>>> data = {"td": 3000}
>>> print(MyModel(**data))
td=datetime.timedelta(seconds=3)