Python 将父类docstring作为_doc__属性继承
有一个关于的问题,但答案涉及方法docstring 我的问题是如何继承父类的docstring作为Python 将父类docstring作为_doc__属性继承,python,docstring,django-rest-framework,Python,Docstring,Django Rest Framework,有一个关于的问题,但答案涉及方法docstring 我的问题是如何继承父类的docstring作为\uuu doc\uu属性。用例是根据视图类的docstring在html版本的API中生成漂亮的文档。但是当在没有docstring的类中继承基类(带有docstring)时,API不会显示docstring class SubClassOverwriteDocstring(ParentWithDocstring): """Original docstring""" def __i
\uuu doc\uu
属性。用例是根据视图类的docstring在html版本的API中生成漂亮的文档。但是当在没有docstring的类中继承基类(带有docstring)时,API不会显示docstring
class SubClassOverwriteDocstring(ParentWithDocstring):
"""Original docstring"""
def __init__(self, docstring, *args, **kwargs):
super(SubClassOverwriteDocstring, self).__init__(*args, **kwargs)
self.docstring = docstring
@property
def __doc__(self):
return self.docstring
>>> subclass = SubClassOverwriteDocstring('new docstring')
>>> print subclass.__doc__ # Prints "new docstring"
new docstring
class SubClassBrokenDocstring(SubClassOverwriteDocstring):
"""Broken docstring"""
def __init__(self, docstring, *args, **kwargs):
super(SubClassBrokenDocstring, self).__init__(docstring, *args, **kwargs)
>>> subclass = SubClassBrokenDocstring("doesn't work")
>>> print subclass.__doc__ # Prints "Broken docstring"
Broken docstring
sphinx和其他工具很可能做了正确的事情,为我处理了docstring继承,但是django rest框架查看了(空)。\uuuuu doc\uuuu
属性
class ParentWithDocstring(object):
"""Parent docstring"""
pass
class SubClassWithoutDoctring(ParentWithDocstring):
pass
parent = ParentWithDocstring()
print parent.__doc__ # Prints "Parent docstring".
subclass = SubClassWithoutDoctring()
print subclass.__doc__ # Prints "None"
我尝试过类似于
super(subclass without docstring,self)。\uuuu doc\uuuu
,但这也只能让我得到None
,因为你不能将一个新的\uuu doc\uuuu
docstring分配给一个类(至少在CPython中),你必须使用一个元类:
import inspect
def inheritdocstring(name, bases, attrs):
if not '__doc__' in attrs:
# create a temporary 'parent' to (greatly) simplify the MRO search
temp = type('temporaryclass', bases, {})
for cls in inspect.getmro(temp):
if cls.__doc__ is not None:
attrs['__doc__'] = cls.__doc__
break
return type(name, bases, attrs)
是的,我们跳过了一个或两个额外的环,但是上面的元类将找到正确的\uuuu doc\uuuu
,不管您的继承图多么复杂
用法:
>>> class ParentWithDocstring(object):
... """Parent docstring"""
...
>>> class SubClassWithoutDocstring(ParentWithDocstring):
... __metaclass__ = inheritdocstring
...
>>> SubClassWithoutDocstring.__doc__
'Parent docstring'
另一种方法是在\uuuuu init\uuuuu
中设置\uuuu doc\uuuuu
,作为实例变量:
def __init__(self):
try:
self.__doc__ = next(cls.__doc__ for cls in inspect.getmro(type(self)) if cls.__doc__ is not None)
except StopIteration:
pass
class ParentWithDocstring(object):
"""Parent docstring"""
pass
class SubClassWithoutDoctring(ParentWithDocstring):
__doc__ = ParentWithDocstring.__doc__
parent = ParentWithDocstring()
print parent.__doc__ # Prints "Parent docstring".
subclass = SubClassWithoutDoctring()
assert subclass.__doc__ == parent.__doc__
那么至少您的实例有一个docstring:
>>> class SubClassWithoutDocstring(ParentWithDocstring):
... def __init__(self):
... try:
... self.__doc__ = next(cls.__doc__ for cls in inspect.getmro(type(self)) if cls.__doc__ is not None)
... except StopIteration:
... pass
...
>>> SubClassWithoutDocstring().__doc__
'Parent docstring'
从Python3.3(已修复)开始,您最终可以设置自定义类的\uuuu doc\uuuu
属性,这样您就可以使用类装饰器了:
import inspect
def inheritdocstring(cls):
for base in inspect.getmro(cls):
if base.__doc__ is not None:
cls.__doc__ = base.__doc__
break
return cls
因此,可以应用:
>>> @inheritdocstring
... class SubClassWithoutDocstring(ParentWithDocstring):
... pass
...
>>> SubClassWithoutDocstring.__doc__
'Parent docstring'
在这种特殊情况下,您还可以通过重写
.get\u name()
方法来重写REST框架如何确定端点使用的名称
如果您真的这样做了,您可能会发现自己想要为视图定义一组基类,并使用一个简单的mixin类覆盖所有基类上的方法
例如:
class GetNameMixin(object):
def get_name(self):
# Your docstring-or-ancestor-docstring code here
class ListAPIView(GetNameMixin, generics.ListAPIView):
pass
class RetrieveAPIView(GetNameMixin, generics.RetrieveAPIView):
pass
还要注意的是,
get\u name
方法被认为是私有的,并且在将来的某个时候可能会更改,因此升级时您需要在发行说明上保留标签,以了解其中的任何更改 最简单的方法是将其指定为类变量:
def __init__(self):
try:
self.__doc__ = next(cls.__doc__ for cls in inspect.getmro(type(self)) if cls.__doc__ is not None)
except StopIteration:
pass
class ParentWithDocstring(object):
"""Parent docstring"""
pass
class SubClassWithoutDoctring(ParentWithDocstring):
__doc__ = ParentWithDocstring.__doc__
parent = ParentWithDocstring()
print parent.__doc__ # Prints "Parent docstring".
subclass = SubClassWithoutDoctring()
assert subclass.__doc__ == parent.__doc__
不幸的是,它是手动的,但很简单。顺便说一句,虽然字符串格式不是通常的工作方式,但它使用相同的方法:
class A(object):
_validTypes = (str, int)
__doc__ = """A accepts the following types: %s""" % str(_validTypes)
A accepts the following types: (<type 'str'>, <type 'int'>)
A类(对象):
_有效类型=(str,int)
__doc_uu=“”A接受以下类型:%s”“%str(_有效类型)
A接受以下类型:(,)
您也可以使用@property
class ParentWithDocstring(object):
"""Parent docstring"""
pass
class SubClassWithoutDocstring(ParentWithDocstring):
@property
def __doc__(self):
return None
class SubClassWithCustomDocstring(ParentWithDocstring):
def __init__(self, docstring, *args, **kwargs):
super(SubClassWithCustomDocstring, self).__init__(*args, **kwargs)
self.docstring = docstring
@property
def __doc__(self):
return self.docstring
>>> parent = ParentWithDocstring()
>>> print parent.__doc__ # Prints "Parent docstring".
Parent docstring
>>> subclass = SubClassWithoutDocstring()
>>> print subclass.__doc__ # Prints "None"
None
>>> subclass = SubClassWithCustomDocstring('foobar')
>>> print subclass.__doc__ # Prints "foobar"
foobar
您甚至可以覆盖docstring
class SubClassOverwriteDocstring(ParentWithDocstring):
"""Original docstring"""
def __init__(self, docstring, *args, **kwargs):
super(SubClassOverwriteDocstring, self).__init__(*args, **kwargs)
self.docstring = docstring
@property
def __doc__(self):
return self.docstring
>>> subclass = SubClassOverwriteDocstring('new docstring')
>>> print subclass.__doc__ # Prints "new docstring"
new docstring
class SubClassBrokenDocstring(SubClassOverwriteDocstring):
"""Broken docstring"""
def __init__(self, docstring, *args, **kwargs):
super(SubClassBrokenDocstring, self).__init__(docstring, *args, **kwargs)
>>> subclass = SubClassBrokenDocstring("doesn't work")
>>> print subclass.__doc__ # Prints "Broken docstring"
Broken docstring
需要注意的是,该属性不能被其他类继承。显然,您必须在要覆盖docstring的每个类中添加该属性
class SubClassOverwriteDocstring(ParentWithDocstring):
"""Original docstring"""
def __init__(self, docstring, *args, **kwargs):
super(SubClassOverwriteDocstring, self).__init__(*args, **kwargs)
self.docstring = docstring
@property
def __doc__(self):
return self.docstring
>>> subclass = SubClassOverwriteDocstring('new docstring')
>>> print subclass.__doc__ # Prints "new docstring"
new docstring
class SubClassBrokenDocstring(SubClassOverwriteDocstring):
"""Broken docstring"""
def __init__(self, docstring, *args, **kwargs):
super(SubClassBrokenDocstring, self).__init__(docstring, *args, **kwargs)
>>> subclass = SubClassBrokenDocstring("doesn't work")
>>> print subclass.__doc__ # Prints "Broken docstring"
Broken docstring
真倒霉!但绝对比做元类的事情容易 我只想从我所记得的关于Python的讨论中补充一下。默认情况下,它不会继承docstring,因为它被认为是因为Python无法知道docstring是否还有意义(尽管继承应该足以意味着程序员遵循正常的OOP,不会完全改变对象的意义),据认为,如果文档为空白,则会减少错误引导。例如,在父类是ABC的情况下,它会变得更加复杂……在3.3中,
\uuuuuu doc\uuuuu
getset\u描述符现在可以为堆类型写入。之前,它只定义了一个getter
;现在它有了设置器
<代码>检查\u设置\u特殊\u类型\u属性
防止删除\uuuuu文档
@eryksun:已确认;这是一个早就应该解决的问题!恢复了我最初仅针对Python 3.3的类装饰器想法。谢谢!最后,我使用了\uuu init\uuu()
一个。元类要求我在很多地方拥有该元类,而不是在一个(基类)类中只有一个init。实际上,我使用了django rest框架的中的init\uuuuuuuuuu()
方法中的inspect.getmro(type(self))
解决方案。您的意思可能是。获取描述()
而不是。get_name()
?是的,我看到了这个,我有一个基类试图覆盖它。但我仍然无法掌握我父母的文档字符串:-)好吧,至少,那是在我读另一个答案之前。“你可能是说。get_description()而不是。get_name()”的确,是的,我做了。看看我最后是怎么做的。注意:根据Martijn的回答(),设置。\uuu doc\uu
只在python 3.3中有效。@ReinoutvanRees,我在2.3.4(是的,两点三)和2.7中试过。在定义类后,不能将指定给._doc,如果为true,则可以在定义时指定。这就是为什么我发布了我的答案。啊,你是对的。因此,如果您记住在定义时实际分配\uuuuuu doc\uuuu
,那么这也是一个选项。很好,很简单。