完整性检查我的SQLAlchemy类和查询是否在功能上等同于原始SQL语句

Question

SQLALchemy新手，这是我尝试重新创建此查询：

SELECT
          database_server_id || 'dr' database_server_id
        , CASE
            WHEN database_server_id LIKE 'cluster1' AND name IS NULL THEN 'canvas'
            WHEN database_server_id LIKE 'cluster4' AND name IS NULL THEN 'canvas'
            WHEN name IS NULL THEN database_server_id
            ELSE name
          END AS database_name
          ,SUBSTRING(database_server_id FROM '\d+$')::INT cluster_id
          ,CASE
            WHEN name IS NULL THEN SUBSTRING(database_server_id FROM '\d+$')::INT
            ELSE SUBSTRING(name FROM '\d+$')::INT
          END AS shard_id
        FROM canvas.switchman_shards
        WHERE database_server_id LIKE 'cluster%'
        ORDER BY cluster_id, shard_id

我使用声明来创建类，然后使用hybrid_properties来处理其他逻辑。让我知道我犯了什么错误！

class SwitchmanShards(Base):
    __tablename__ = 'switchman_shards'

    id = Column(Integer, primary_key=True)
    database_server_id = Column(String)
    name = Column(String, nullable=True)
    default = Column(Boolean)
    settings = Column(String)
    delayed_jobs_shard_id = Column(Integer, nullable=True)

    def __init__(self,
                 id,
                 db_server_id,
                 name,
                 default,
                 settings,
                 del_job_shard_id):
        self.id = id
        self.database_server_id = db_server_id
        self.name = name
        self.default = default
        self.settings = settings
        self.delayed_jobs_shard_id = del_job_shard_id

    @hybrid_property
    def cluster_id(self):
        # Clusters that do not use their cluster name as the schema and need to
        # be recast.
        cluster_list = ('cluster1dr', 'cluster4dr')
        if self.database_server_id in cluster_list and self.name is None:
            return 'canvas'
        elif self.name is None:
            return self.database_server_id
        else:
            return self.name

    @hybrid_property
    def shard_id(self):
        shard_regex = re.compile(r'(\d+$)')
        if self.name is None:
            res = shard_regex.search(self.database_server_id)
            return res.group(1)
        else:
            res = shard_regex.search(self.name)
            return res.group(1)



def return_shards():
    Session = sessionmaker(bind=engine)
    session = Session()
    shards = session.query(SwitchmanShards).filter(
        SwitchmanShards.database_server_id.like('cluster%')
    ).all()
    return shards

作为旁注，我试图弄清楚SQLALchemy的价值，因为我花了大约3倍的行来编写相同的代码。可以说它更具可读性，这是我能看到的最大优点。

编辑：我注意到这比原始SQL查询少了10个记录，不知道为什么会这样。