CODEWITHSUNDEEP
pythonpython-3.xpython-dataclasses
user3310334
user3310334

Reputation:

Customise dataclass asdict keynames

How can I create a dataclass that has a custom dict representation?

Example dict output:

{
    '$1': 'foo',
    '$2': 'bar',
    '$3': 'baz'
}

Obviously I can't have dataclass

@dataclass
class Foo:
    $1: str
    ...

I would need it represented as

@dataclass
class Foo:
    one: str
    ...

for example, but I would like dataclasses.asdict(foo) to return with the "$1" etc. key names.

Also it would be great if

x = {
    '$1': 'foo',
    '$2': 'bar',
    '$3': 'baz'
}
foo = Foo(**x)
assert foo.one == 'foo'
assert foo.two == 'bar'
assert foo.three == 'baz'

worked

Upvotes: 3

Views: 4144

Answers (1)

chepner
chepner

Reputation: 532003

You can simply pass your own dict_factory argument, which uses your mapping to construct the dict.

@dataclass
class Foo:
    one: str

def factory(kv_pairs):
    m = {
        'one': '$1',
        # etc
    }
    return {m.get(k): v for k, v in kv_pairs}


f = Foo("foo")
assert dataclasses.asdict(f, dict_factory=factory) == {'$1': 'foo'}

For initializing the class, I would go with a custom class method.

m = {'one': '$1'}
m_inv = {v: k for k, v in m.items()}

@dataclass
class Foo:
    one: str

    @classmethod
    def from_dollar_vars(cls, **kwargs):
        return cls(**{m_inv[k]: v for k, v in kv_pairs})

def factory(kv_pairs):
    return {m[k]: v for k, v in kv_pairs}

x = {'$1': 'foo'}
f = Foo.from_dollar_vars(**x)
assert dataclasses.asdict(f, dict_factory=factory) == x

Upvotes: 6

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