Set initial value of a Pydantic 2 field using validator

Question

Say I have the following Pydantic 2.10.6 model, where x is dynamically calculated from another field, y:

from pydantic import BaseModel, Field, field_validator, ValidationInfo

class Foo(BaseModel):
    x: int = Field(init=False)
    y: int

    @field_validator("x", mode="before")
    @staticmethod
    def set_x(val: None, info: ValidationInfo):
        return info.data["y"] + 1

Foo(y=1)

Running this as is fails validation:

pydantic_core._pydantic_core.ValidationError: 1 validation error for Foo
x
  Field required [type=missing, input_value={'y': 1}, input_type=dict]

I can "fix" the runtime error by giving x a default:

    x: int = Field(init=False, default=None)

But then this fails type checking in pyright with:

Type "None" is not assignable to declared type "int"

I can also fix it using a model_validator, but this is a tad less neat:

from pydantic import BaseModel, Field, model_validator

class Foo(BaseModel):
    x: int = Field(init=False)
    y: int

    @model_validator(mode="before")
    @staticmethod
    def set_x(data: dict):
        data["x"] = data["y"] + 1
        return data

Foo(y=1)

How can I cleanly represent this model using only a field validator?

Does x have to be a regular field then? If you are initializing it from y, why not using a computed_field? — lord_haffi
– lord_haffi, Commented Mar 21 at 8:49

Marco Bresson · Accepted Answer · 2025-04-15 12:03:53Z

1

as @lord_haffi said, you can use computed fields in pydantic. Given your example, you would have:

from pydantic import BaseModel, computed_field

class Foo(BaseModel):
    y: int

    @computed_field
    @property
    def x(self) -> int:
        return self.y + 1

Foo(y=1)
# Foo(y=1, x=2)

answered Apr 15 at 12:03

Marco Bresson

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