Python의 클래스 인스턴스에서 어떤 기능을 사용할 수 있습니까?
-
11-09-2019 - |
문제
클래스 인스턴스에서 어떤 기능이 정의되었는지 어떻게 동적으로 찾으십니까?
예를 들어:
class A(object):
def methodA(self, intA=1):
pass
def methodB(self, strB):
pass
a = A()
이상적으로는 'a'인스턴스에 Methoda와 Methodb가 있고 어떤 주장이 취해야한다는 것을 알고 싶습니다.
해결책
살펴보십시오 inspect
기준 치수.
>>> import inspect
>>> inspect.getmembers(a)
[('__class__', <class '__main__.A'>),
('__delattr__', <method-wrapper '__delattr__' of A object at 0xb77d48ac>),
('__dict__', {}),
('__doc__', None),
('__getattribute__',
<method-wrapper '__getattribute__' of A object at 0xb77d48ac>),
('__hash__', <method-wrapper '__hash__' of A object at 0xb77d48ac>),
('__init__', <method-wrapper '__init__' of A object at 0xb77d48ac>),
('__module__', '__main__'),
('__new__', <built-in method __new__ of type object at 0x8146220>),
('__reduce__', <built-in method __reduce__ of A object at 0xb77d48ac>),
('__reduce_ex__', <built-in method __reduce_ex__ of A object at 0xb77d48ac>),
('__repr__', <method-wrapper '__repr__' of A object at 0xb77d48ac>),
('__setattr__', <method-wrapper '__setattr__' of A object at 0xb77d48ac>),
('__str__', <method-wrapper '__str__' of A object at 0xb77d48ac>),
('__weakref__', None),
('methodA', <bound method A.methodA of <__main__.A object at 0xb77d48ac>>),
('methodB', <bound method A.methodB of <__main__.A object at 0xb77d48ac>>)]
>>> inspect.getargspec(a.methodA)
(['self', 'intA'], None, None, (1,))
>>> inspect.getargspec(getattr(a, 'methodA'))
(['self', 'intA'], None, None, (1,))
>>> print inspect.getargspec.__doc__
Get the names and default values of a function's arguments.
A tuple of four things is returned: (args, varargs, varkw, defaults).
'args' is a list of the argument names (it may contain nested lists).
'varargs' and 'varkw' are the names of the * and ** arguments or None.
'defaults' is an n-tuple of the default values of the last n arguments.
>>> print inspect.getmembers.__doc__
Return all members of an object as (name, value) pairs sorted by name.
Optionally, only return members that satisfy a given predicate.
제휴하지 않습니다 StackOverflow