返回this:class:的内容。KeyedTuple作为字典

In [46]: result = aggregate_events[0]

In [47]: type(result)
Out[47]: sqlalchemy.util._collections.result

In [48]: def to_dict(query_result=None):
    ...:     cover_dict = {key: getattr(query_result, key) for key in query_result.keys()}
    ...:     return cover_dict
    ...: 
    ...:     

In [49]: to_dict(result)
Out[49]: 
{'calculate_avg': None,
 'calculate_max': None,
 'calculate_min': None,
 'calculate_sum': None,
 'dataPointIntID': 6,
 'data_avg': 10.0,
 'data_max': 10.0,
 'data_min': 10.0,
 'data_sum': 60.0,
 'deviceID': u'asas',
 'productID': u'U7qUDa',
 'tenantID': u'CvdQcYzUM'}

为了完成@Anurag Uniyal的回答，这里有一个递归地遵循关系的方法:

from sqlalchemy.inspection import inspect

def to_dict(obj, with_relationships=True):
    d = {}
    for column in obj.__table__.columns:
        if with_relationships and len(column.foreign_keys) > 0:
             # Skip foreign keys
            continue
        d[column.name] = getattr(obj, column.name)

    if with_relationships:
        for relationship in inspect(type(obj)).relationships:
            val = getattr(obj, relationship.key)
            d[relationship.key] = to_dict(val) if val else None
    return d

class User(Base):
    __tablename__ = 'users'
    id = Column(Integer, primary_key=True)
    first_name = Column(TEXT)
    address_id = Column(Integer, ForeignKey('addresses.id')
    address = relationship('Address')

class Address(Base):
    __tablename__ = 'addresses'
    id = Column(Integer, primary_key=True)
    city = Column(TEXT)


user = User(first_name='Nathan', address=Address(city='Lyon'))
# Add and commit user to session to create ids

to_dict(user)
# {'id': 1, 'first_name': 'Nathan', 'address': {'city': 'Lyon'}}
to_dict(user, with_relationship=False)
# {'id': 1, 'first_name': 'Nathan', 'address_id': 1}

有了这段代码，您还可以添加到您的查询“过滤器”或“连接”，这工作!

query = session.query(User)
def query_to_dict(query):
        def _create_dict(r):
            return {c.get('name'): getattr(r, c.get('name')) for c in query.column_descriptions}

    return [_create_dict(r) for r in query]

在大多数情况下，列名适合它们。但是你可能会像下面这样写代码:

class UserModel(BaseModel):
    user_id = Column("user_id", INT, primary_key=True)
    email = Column("user_email", STRING)

column.name“user_email”而字段名是“email”，column.name不能像以前那样工作。

sqlalchemy_base_model.py

我把答案写在这里

老问题，但由于这是谷歌中“sqlalchemy row to dict”的第一个结果，它值得一个更好的答案。

SqlAlchemy返回的RowProxy对象具有items()方法: http://docs.sqlalchemy.org/en/latest/core/connections.html#sqlalchemy.engine.RowProxy.items

它只是返回一个(key, value)元组列表。因此可以使用以下方法将行转换为dict:

在Python中<= 2.6:

rows = conn.execute(query)
list_of_dicts = [dict((key, value) for key, value in row.items()) for row in rows]

在Python中>= 2.7:

rows = conn.execute(query)
list_of_dicts = [{key: value for (key, value) in row.items()} for row in rows]

在SQLAlchemy v0.8及更新版本中，使用检查系统。

from sqlalchemy import inspect

def object_as_dict(obj):
    return {c.key: getattr(obj, c.key)
            for c in inspect(obj).mapper.column_attrs}

user = session.query(User).first()

d = object_as_dict(user)

注意.key是属性名，可以与列名不同，例如:

class_ = Column('class', Text)

此方法也适用于column_property。