python - 如何将我自己的 "help"信息添加到 python 函数/类中？

Question

我知道我可以使用“help()”来查看包中现有的帮助信息。但是在我写了自己的函数/类之后，我怎样才能启用“帮助”来查看帮助文档呢？我知道“comment”的第一行是 doc 属性，但这不是我想要的。

我希望编译出自己的包，其他人可以从"help()"中看到。如何做到这一点？

Answer 1

help() 完全基于 __doc__ 属性(以及函数参数的内省(introspection))，因此请确保您的模块、类和函数都有文档字符串。

文档字符串不是注释，它是顶部的裸字符串文字:

"""This is a module docstring, shown when you use help() on a module"""

class Foo:
    """Help for the class Foo"""

    def bar(self):
        """Help for the bar method of Foo classes"""

def spam(f):
    """Help for the spam function"""

例如，流行的第三方requests模块有一个docstring:

>>> import requests
>>> requests.__doc__
'\nRequests HTTP library\n~~~~~~~~~~~~~~~~~~~~~\n\nRequests is an HTTP library, written in Python, for human beings. Basic GET\nusage:\n\n   >>> import requests\n   >>> r = requests.get(\'https://www.python.org\')\n   >>> r.status_code\n   200\n   >>> \'Python is a programming language\' in r.content\n   True\n\n... or POST:\n\n   >>> payload = dict(key1=\'value1\', key2=\'value2\')\n   >>> r = requests.post(\'http://httpbin.org/post\', data=payload)\n   >>> print(r.text)\n   {\n     ...\n     "form": {\n       "key2": "value2",\n       "key1": "value1"\n     },\n     ...\n   }\n\nThe other HTTP methods are supported - see `requests.api`. Full documentation\nis at <http://python-requests.org>.\n\n:copyright: (c) 2016 by Kenneth Reitz.\n:license: Apache 2.0, see LICENSE for more details.\n'

由 help() 直接呈现，连同模块内容(以及递归地，内容文档字符串):

>>> help('requests')
Help on package requests:

NAME
    requests

DESCRIPTION
    Requests HTTP library
    ~~~~~~~~~~~~~~~~~~~~~

    Requests is an HTTP library, written in Python, for human beings. Basic GET
    usage:

       >>> import requests
       >>> r = requests.get('https://www.python.org')
       >>> r.status_code
       200
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