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问题:
I want to be able to create a python decorator that automatically "registers" class methods in a global repository (with some properties).
Example code:
class my_class(object):
@register(prop1,prop2)
def my_method( arg1,arg2 ):
# method code here...
@register(prop3,prop4)
def my_other_method( arg1,arg2 ):
# method code here...
I want that when loading is done, somewhere there will be a dict containing:
{ "my_class.my_method" : ( prop1, prop2 )
"my_class.my_other_method" : ( prop3, prop4 ) }
Is this possible?
回答1:
Not with just a decorator, no. But a metaclass can automatically work with a class after its been created. If your register
decorator just makes notes about what the metaclass should do, you can do the following:
registry = {}
class RegisteringType(type):
def __init__(cls, name, bases, attrs):
for key, val in attrs.iteritems():
properties = getattr(val, 'register', None)
if properties is not None:
registry['%s.%s' % (name, key)] = properties
def register(*args):
def decorator(f):
f.register = tuple(args)
return f
return decorator
class MyClass(object):
__metaclass__ = RegisteringType
@register('prop1','prop2')
def my_method( arg1,arg2 ):
pass
@register('prop3','prop4')
def my_other_method( arg1,arg2 ):
pass
print registry
printing
{'MyClass.my_other_method': ('prop3', 'prop4'), 'MyClass.my_method': ('prop1', 'prop2')}
回答2:
Here's a little love for class decorators. I think the syntax is slightly simpler than that required for metaclasses.
def class_register(cls):
cls._propdict = {}
for methodname in dir(cls):
method = getattr(cls, methodname)
if hasattr(method, '_prop'):
cls._propdict.update(
{cls.__name__ + '.' + methodname: method._prop})
return cls
def register(*args):
def wrapper(func):
func._prop = args
return func
return wrapper
@class_register
class MyClass(object):
@register('prop1', 'prop2')
def my_method(self, arg1, arg2):
pass
@register('prop3', 'prop4')
def my_other_method(self, arg1, arg2):
pass
myclass = MyClass()
print(myclass._propdict)
# {'MyClass.my_other_method': ('prop3', 'prop4'), 'MyClass.my_method': ('prop1', 'prop2')}
回答3:
If you need the classes name, use Matt's solution. However, if you're ok with just having the methods name -- or a reference to the method -- in the registry, this might be a simpler way of doing it:
class Registry:
r = {}
@classmethod
def register(cls, *args):
def decorator(fn):
cls.r[fn.__name__] = args
return fn
return decorator
class MyClass(object):
@Registry.register("prop1","prop2")
def my_method( arg1,arg2 ):
pass
@Registry.register("prop3","prop4")
def my_other_method( arg1,arg2 ):
pass
print Registry.r
print
{'my_other_method': ('prop3', 'prop4'), 'my_method': ('prop1', 'prop2')}
回答4:
Not as beautiful or elegant, but probably the simplest way if you only need this in one class only:
_registry = {}
class MyClass(object):
def register(*prop):
def decorator(meth):
_registry[MyClass.__name__ + '.' + meth.__name__] = prop
return decorator
@register('prop1', 'prop2')
def my_method(self, arg1, arg2):
pass
@register('prop3', 'prop4')
def my_other_method(self, arg1, arg2):
pass
del register
回答5:
Not easy, but if you are using Python 3 this should work:
registry = {}
class MetaRegistry(type):
@classmethod
def __prepare__(mcl, name, bases):
def register(*props):
def deco(f):
registry[name + "." + f.__name__] = props
return f
return deco
d = dict()
d['register'] = register
return d
def __new__(mcl, name, bases, dct):
del dct['register']
cls = super().__new__(mcl, name, bases, dct)
return cls
class my_class(object, metaclass=MetaRegistry):
@register('prop1','prop2')
def my_method( arg1,arg2 ):
pass # method code here...
@register('prop3','prop4')
def my_other_method( arg1,arg2 ):
pass # method code here...
print(registry)
Note that you can not have method names equal to the decorator name in the meta-typed class, because they are automatically deleted by the del
command in the metaclass's __new__
method.
For Python 2.6 I think you would have to explicitly tell the decorator the class name to use.
回答6:
No. The decorator receives the function before it has become a method, so you don't know what class it is on.