我需要一种工作方法来获取从Python基类继承的所有类。
当前回答
下面是一个没有递归的版本:
def get_subclasses_gen(cls):
def _subclasses(classes, seen):
while True:
subclasses = sum((x.__subclasses__() for x in classes), [])
yield from classes
yield from seen
found = []
if not subclasses:
return
classes = subclasses
seen = found
return _subclasses([cls], [])
这与其他实现的不同之处在于它返回原始类。 这是因为它使代码更简单,并且:
class Ham(object):
pass
assert(issubclass(Ham, Ham)) # True
如果get_subclasses_gen看起来有点奇怪,那是因为它是通过将尾递归实现转换为循环生成器创建的:
def get_subclasses(cls):
def _subclasses(classes, seen):
subclasses = sum(*(frozenset(x.__subclasses__()) for x in classes))
found = classes + seen
if not subclasses:
return found
return _subclasses(subclasses, found)
return _subclasses([cls], [])
其他回答
如果你只想要直接的子类,那么.__subclasses__()就可以了。如果你想要所有的子类,子类的子类等等,你需要一个函数来为你做这些。
下面是一个简单易读的函数,它可以递归地找到给定类的所有子类:
def get_all_subclasses(cls):
all_subclasses = []
for subclass in cls.__subclasses__():
all_subclasses.append(subclass)
all_subclasses.extend(get_all_subclasses(subclass))
return all_subclasses
这个答案不如使用@unutbu提到的特殊内置__subclasses__()类方法好,所以我只是把它作为一个练习。subclasses()函数的定义返回一个字典,该字典将所有子类名称映射到子类本身。
def traced_subclass(baseclass):
class _SubclassTracer(type):
def __new__(cls, classname, bases, classdict):
obj = type(classname, bases, classdict)
if baseclass in bases: # sanity check
attrname = '_%s__derived' % baseclass.__name__
derived = getattr(baseclass, attrname, {})
derived.update( {classname:obj} )
setattr(baseclass, attrname, derived)
return obj
return _SubclassTracer
def subclasses(baseclass):
attrname = '_%s__derived' % baseclass.__name__
return getattr(baseclass, attrname, None)
class BaseClass(object):
pass
class SubclassA(BaseClass):
__metaclass__ = traced_subclass(BaseClass)
class SubclassB(BaseClass):
__metaclass__ = traced_subclass(BaseClass)
print subclasses(BaseClass)
输出:
{'SubclassB': <class '__main__.SubclassB'>,
'SubclassA': <class '__main__.SubclassA'>}
虽然我非常倾向于__init_subclass__方法,这将保留定义顺序,并避免组合增长顺序,如果你有一个非常密集的层次结构,到处都有多个继承:
def descendents(cls):
'''Does not return the class itself'''
R = {}
def visit(cls):
for subCls in cls.__subclasses__():
R[subCls] = True
visit(subCls)
visit(cls)
return list(R.keys())
这是因为字典会记住键的插入顺序。列表方法也会起作用。
我怎么能找到一个类的所有子类给它的名字?
我们当然可以很容易地做到这一点,只要能访问对象本身。
仅仅给出它的名字是一个糟糕的想法,因为可以有多个同名的类,甚至在同一个模块中定义。
我为另一个答案创建了一个实现,因为它回答了这个问题,而且它比这里的其他解决方案更优雅,下面是:
def get_subclasses(cls):
"""returns all subclasses of argument, cls"""
if issubclass(cls, type):
subclasses = cls.__subclasses__(cls)
else:
subclasses = cls.__subclasses__()
for subclass in subclasses:
subclasses.extend(get_subclasses(subclass))
return subclasses
用法:
>>> import pprint
>>> list_of_classes = get_subclasses(int)
>>> pprint.pprint(list_of_classes)
[<class 'bool'>,
<enum 'IntEnum'>,
<enum 'IntFlag'>,
<class 'sre_constants._NamedIntConstant'>,
<class 'subprocess.Handle'>,
<enum '_ParameterKind'>,
<enum 'Signals'>,
<enum 'Handlers'>,
<enum 'RegexFlag'>]
下面是一个没有递归的版本:
def get_subclasses_gen(cls):
def _subclasses(classes, seen):
while True:
subclasses = sum((x.__subclasses__() for x in classes), [])
yield from classes
yield from seen
found = []
if not subclasses:
return
classes = subclasses
seen = found
return _subclasses([cls], [])
这与其他实现的不同之处在于它返回原始类。 这是因为它使代码更简单,并且:
class Ham(object):
pass
assert(issubclass(Ham, Ham)) # True
如果get_subclasses_gen看起来有点奇怪,那是因为它是通过将尾递归实现转换为循环生成器创建的:
def get_subclasses(cls):
def _subclasses(classes, seen):
subclasses = sum(*(frozenset(x.__subclasses__()) for x in classes))
found = classes + seen
if not subclasses:
return found
return _subclasses(subclasses, found)
return _subclasses([cls], [])
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