为什么我要在Python中设置部分对象的_doc___;?
指部分物体。 然而,这本书中的例子说明了这一点:为什么我要在Python中设置部分对象的_doc___;?,python,functools,Python,Functools,指部分物体。 然而,这本书中的例子说明了这一点: 如果没有帮助,为什么要设置\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu def somefunc(somerandomarg): """This is a docstring What I am doing is shorthand for __doc__ = "
如果没有帮助,为什么要设置
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
def somefunc(somerandomarg):
"""This is a docstring
What I am doing is shorthand for __doc__ = "random docstring"
For partials, you can't do this, so you set it with __doc__"""
pass
记录任何东西都是好的,即使是部分内容。显然,pydoc
不知道partial
对象,也不希望它们有修改。理论上,更智能的工具也会遇到同样的问题。我在这里演示了如何重写partial()
,作为一个保留\uuuu doc\uuuu
的装饰器工厂:我希望它们也是一样的。但是如果你阅读我链接的help()
问题,你会发现它们是不同的;分部不是函数,而是对象help()
将显示分部类及其方法的文档。正如在链接中所说的,如果不修改Python代码,就无法覆盖它。但是,在Python中,当您键入docstring时,pydoc将显示它。键入它是什么意思?当我在一个part.py中键入这个时,pydoc part.basetwo
显示partial的文档,而不是我键入的文档。进入Python Shell,在文档中定义一个类似于basetwo的分部函数。然后输入basetwo(
)。您的文档将显示。我认为帮助不起作用的原因是没有人对pydoc进行编码以了解部分对象的功能性质。但是,都可以访问直接basetwo.\uuuu doc\uuuuu
并键入basetwo(
会起作用,所以在某些情况下设置\uuu doc\uuu
实际上是有帮助的。
def somefunc(somerandomarg):
"""This is a docstring
What I am doing is shorthand for __doc__ = "random docstring"
For partials, you can't do this, so you set it with __doc__"""
pass