导入处理/模块从python2更改为python3？

我试图遵循 this previous question中与SQLAlchemy相关的设计模式,并打算在多个文件之间共享一个公共Base实例.完全相同的代码适用于python2和python3.

但是,当我在一个模块(称为模型)中移动文件a.py,b.py,c.py和base.py并添加必要的__init__.py文件时,它继续在python2上工作,但随后产生错误在python3上(详情如下).

我有以下文件：

模型/ base.py

from sqlalchemy.ext.declarative import declarative_base
Base = declarative_base()

模型/ a.py

from sqlalchemy import *
from base import Base
from sqlalchemy.orm import relationship

class A(Base):
    __tablename__ = "A"
    id  = Column(Integer,primary_key=True)
    Bs  = relationship("B",backref="A.id")
    Cs  = relationship("C",backref="A.id")

模型/ b.py

from sqlalchemy import *
from base import Base

class B(Base):
    __tablename__ = "B"
    id    = Column(Integer,primary_key=True)
    A_id  = Column(Integer,ForeignKey("A.id"))

模型/ c.py

from sqlalchemy import *
from base import Base

class C(Base):
    __tablename__ = "C"    
    id    = Column(Integer,ForeignKey("A.id"))

模型/ __ init__.py

(空)

main.py

from sqlalchemy import create_engine
from sqlalchemy.orm import relationship,backref,sessionmaker

from model import base


from model import a
from model import b
from model import c

engine = create_engine("sqlite:///:memory:")
base.Base.metadata.create_all(engine,checkfirst=True)
Session = sessionmaker(bind=engine)
session = Session()

a1 = a.A()
b1 = b.B()
b2 = b.B()
c1 = c.C()
c2 = c.C()

a1.Bs.append(b1)
a1.Bs.append(b2)    
a1.Cs.append(c1)
a1.Cs.append(c2)    
session.add(a1)
session.commit()

python2的工作原理：

$python main.py ; echo $?
0

python3错误：

$python3 main.py ; echo $?
Traceback (most recent call last):
  File "main.py",line 7,in <module>
    from model import a
  File "/home/shale/code/py/try/model/a.py",line 2,in <module>
    from base import Base
ImportError: No module named base
1

我最终通过将base.py中的代码放入我的__init__.py文件中解决了这个问题(如下面的一个答案所述),但有没有人知道为什么这会在python3中产生错误但在python2中不会产生错误？首先是什么变化对此负责？

from .base import Base

from __future__ import absolute_import