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Tishka17 avatar Tishka17 commented on August 15, 2024

Как вариант:
Если name_mapping - список, а не словарь, дампить в список.
При этом все неуказанные поля добавлять в порядке их следования в исходном классе.

@dataclass
class A:
    first: int
    second: int

Если взять схему:

schema = Schema(
    name_mapping=[]
)

Превратит объект A(1,2) в [1,2]

Если взять схему:

schema = Schema(
    name_mapping=['second']
)

Превратит объект A(1,2) в [2,1]

Если взять схему:

schema = Schema(
    name_mapping=['second'],
    only_mapped=True,
)

Превратит объект A(1,2) в [2]

Проблема: нельзя сделать иерархию внутри списка как в случае с маппингом. Но возможно это и не нужно.

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Tishka17 avatar Tishka17 commented on August 15, 2024

В текущий момент возможно сделать список из датакласса указав ВСЕ необходимые поля в name_mapping в виде

schema = Schema(name_mapping={
  "first": (0, ),
  "second": (1, )
})

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zhPavel avatar zhPavel commented on August 15, 2024

Now we have mapping to list

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