[python-flask] from_dict(to_dict(obj)) does not work (snake_case vs CamelCase)

Description

When converting a model to and from a dict, the following command statement does not work.
It removes all the fields being defined as CamelCase in swagger but implemented as snake_case in python.

settings.from_dict(settings.to_dict(obj))

Swagger-codegen version

2.3.1

Swagger declaration file content or url

...
definitions:
  Settings:
    type: object
    properties:
      someArg:
        type: string
...

Command line used for generation

java -jar swagger-codegen-cli-2.3.1.jar generate -i swagger.yml -l python-flask -o flask

Steps to reproduce

settings = Settings(some_args='something')
other_settings = settings.from_dict(settings.to_dict())
# -> other_settings.some_args is None

Related issues/PRs

• I will create a PR to solve this issue if my fix is accepted by the community

Suggest a fix/enhancement

Replace swagger_server.util.deserialize_model(.) with the following version:

def deserialize_model(data, klass):
    """Deserializes list or dict to model.

    :param data: dict, list.
    :type data: dict | list
    :param klass: class literal.
    :return: model object.
    """
    instance = klass()

    if not instance.swagger_types:
        return data

    for attr, attr_type in six.iteritems(instance.swagger_types):
        if data is not None and isinstance(data, (list, dict)):
            value = None
            if instance.attribute_map[attr] in data:
                value = data[instance.attribute_map[attr]]
            elif attr in data:
                value = data[attr]

            setattr(instance, attr, _deserialize(value, attr_type))

    return instance

Locate the Python-Flask template implementation of swagger_server.util.deserialize_model and compare it with the proposed version in the issue. Reproduce the supplied Settings round trip from_dict(to_dict(obj)) and verify that the some_args value survives when the Swagger property is someArg.