python中的简单交叉导入
我想在不同的类中分离代码,并将它们放在不同的文件中。然而,这些类是相互依赖的 main.py:python中的简单交叉导入,python,import,cross-reference,Python,Import,Cross Reference,我想在不同的类中分离代码,并将它们放在不同的文件中。然而,这些类是相互依赖的 main.py: from lib import A, B def main(): a = A() b = B() a.hello() b.hello() if __name__ == '__main__': main() import lib.B class A(): def __init__(self): print "A" def
from lib import A, B
def main():
a = A()
b = B()
a.hello()
b.hello()
if __name__ == '__main__':
main()
import lib.B
class A():
def __init__(self):
print "A"
def hello(self):
print "hello A"
b = B()
import lib.A
class B():
def __init__(self):
print "B"
def hello(self):
print "hello B"
a = A()
lib/_init.py:
from a import A
from b import B
lib/a.py:
from lib import A, B
def main():
a = A()
b = B()
a.hello()
b.hello()
if __name__ == '__main__':
main()
import lib.B
class A():
def __init__(self):
print "A"
def hello(self):
print "hello A"
b = B()
import lib.A
class B():
def __init__(self):
print "B"
def hello(self):
print "hello B"
a = A()
lib/b.py:
from lib import A, B
def main():
a = A()
b = B()
a.hello()
b.hello()
if __name__ == '__main__':
main()
import lib.B
class A():
def __init__(self):
print "A"
def hello(self):
print "hello A"
b = B()
import lib.A
class B():
def __init__(self):
print "B"
def hello(self):
print "hello B"
a = A()
在Python中可以这样做吗?
编辑:
from lib import A, B
def main():
a = A()
b = B()
a.hello()
b.hello()
if __name__ == '__main__':
main()
import lib.B
class A():
def __init__(self):
print "A"
def hello(self):
print "hello A"
b = B()
import lib.A
class B():
def __init__(self):
print "B"
def hello(self):
print "hello B"
a = A()
我收到以下错误消息:
pydev debugger: starting
Traceback (most recent call last):
File "eclipse-python/plugins/org.python.pydev_2.7.1.2012100913/pysrc/pydevd.py", line 1397, in <module>
debugger.run(setup['file'], None, None)
File "eclipse-python/plugins/org.python.pydev_2.7.1.2012100913/pysrc/pydevd.py", line 1090, in run
pydev_imports.execfile(file, globals, locals) #execute the script
File "main.py", line 2, in <module>
from lib import A, B
File "lib/__init__.py", line 1, in <module>
from a import A
File "lib/a.py", line 1, in <module>
import lib.B
ImportError: No module named B
pydev调试器:正在启动
回溯(最近一次呼叫最后一次):
文件“eclipsepython/plugins/org.python.pydev_2.7.1.2012100913/pysrc/pydevd.py”,第1397行,在
运行(安装程序['file'],无,无)
文件“eclipsepython/plugins/org.python.pydev_2.7.1.2012100913/pysrc/pydevd.py”,第1090行,正在运行
pydev_imports.execfile(文件、全局、局部)#执行脚本
文件“main.py”,第2行,在
从库导入A、B
文件“lib/\uuuu init\uuuuu.py”,第1行,在
从进口
文件“lib/a.py”,第1行,在
导入lib.B
ImportError:没有名为B的模块
您可以在hello函数中导入另一个模块,而不是导入顶部的模块
class B():
def __init__(self):
print "B"
def hello(self):
from lib import A
print "hello B"
a = A()
如果只想导入一次,可以在类的构造函数中导入,并使变量为全局变量:
class B():
def __init__(self):
global A
from lib import A
print "B"
def hello(self):
print "hello B"
a = A()
这将把一个全局变量导入到一个全局变量中,并使其可以在模块内访问。当您有两个相互依赖的类时,通常意味着它们实际上属于同一个模块,或者您的耦合太紧,应该使用依赖项注入来解决
现在确实有两种情况,从函数内部导入是“最差”的解决方案,但这仍然是您应该尽量避免的。您的主要问题是,您试图导入一个类,但使用的语法只能导入一个模块。特别是,如果
A
是在模块lib.A
中定义的类(并导入到lib
的顶级命名空间中),那么import lib.A
将永远无法工作
我建议您避免使用from\uimport
语法,除非您确实需要它。这使得依赖关系更容易解决:
lib/a.py
:
import lib.b # note, we're not importing the class B, just the module b!
class A():
def foo(self):
return lib.b.B() # use the class later, with a qualified name
import lib.a # again, just import the module, not the class
class B():
def foo(self):
return lib.a.A() # use another qualified name reference
from a import A # these imports are fine, since the sub-modules don't rely on them
from b import B # they can be the public API for the A and B classes
lib/b.py
:
import lib.b # note, we're not importing the class B, just the module b!
class A():
def foo(self):
return lib.b.B() # use the class later, with a qualified name
import lib.a # again, just import the module, not the class
class B():
def foo(self):
return lib.a.A() # use another qualified name reference
from a import A # these imports are fine, since the sub-modules don't rely on them
from b import B # they can be the public API for the A and B classes
lib/\uuuuu init\uuuuu.py
:
import lib.b # note, we're not importing the class B, just the module b!
class A():
def foo(self):
return lib.b.B() # use the class later, with a qualified name
import lib.a # again, just import the module, not the class
class B():
def foo(self):
return lib.a.A() # use another qualified name reference
from a import A # these imports are fine, since the sub-modules don't rely on them
from b import B # they can be the public API for the A and B classes
如果不希望a
和b
依赖于其包lib
的名称,也可以使用相对模块导入
这是肯定会起作用的,因为A类或B类实际上都不需要存在另一类才能定义。只有在它们被导入之后,A的实例才需要了解B类(反之亦然)
如果其中一个类继承自另一个类,或者以其他方式在顶层使用另一个类的实例,那么您需要更加小心先加载的是哪个模块,否则它可能仍然会损坏。您尝试过吗?出了什么问题?可以对整个类进行这样的导入,或者我在需要lib.A的每个方法中都这样做?您可以在顶部使用
import lib
,然后使用lib.A()
如果键入太多,请尝试import lib as x
并使用x.A()
更好地使用from lib import A
…