argparse：如何调用方法而不是函数？

发布于 2024-10-11 01:15:51 字数 333 浏览 0 评论 0原文

我在 python 程序中使用 argparse，当我运行我的软件时，我想调用一个方法，如下所示：

$ python __init__.py foo

如果我想调用一个函数而不是一个方法，这很容易：

def foo(args):
    pass

def main():
    foo_parser.set_defaults(func=foo)

if __name__ == "__main__":
    main()

我怎样才能替换“ func=foo”通过“func=MyClass.my_method”？

原文

I am using argparse in my python program and I want to call a method when I run my software like this :

$ python __init__.py foo

It is easy if I want to call a function instead of a method :

def foo(args):
    pass

def main():
    foo_parser.set_defaults(func=foo)

if __name__ == "__main__":
    main()

How can I replace "func=foo" by "func=MyClass.my_method" ?

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

与风相奔跑 2024-10-18 01:15:51

您只需引用方法而不是函数。是的。就是这么简单。而不是

 func = foo

You do

 func = theobject.foo

Done!

You just refer to the method instead of the function. Yup. It's that easy. Instead of

 func = foo

You do

 func = theobject.foo

Done!

回复收藏 0 原文

一曲琵琶半遮面シ 2024-10-18 01:15:51

你自己给出了正确答案。只需输入：

func = Your_Class.your_method

而不是：

func = foo

You gave the correct answer yourself. Just put:

func = Your_Class.your_method

instead of:

func = foo

回复收藏 0 原文

揽月 2024-10-18 01:15:51

您是否正在寻找这样的东西：

import argparse

class Myclass(object):

    def foo(self):
        print 'foo'

    def bar(self):
        print 'bar'

class Main(Myclass):

    def __init__(self):
        foo_parser = argparse.ArgumentParser()

        foo_parser.add_argument('method')
        self.args = foo_parser.parse_args()

    def __call__(self, *args, **kws):
        method = self.args.method
        return getattr(self, method)(*args, **kws)

if __name__ == "__main__":
    main = Main()

    main()

用法示例：

$ python test.py foo
'foo'
$ python test.py bar
'bar'

Are you looking for something like this :

import argparse

class Myclass(object):

    def foo(self):
        print 'foo'

    def bar(self):
        print 'bar'

class Main(Myclass):

    def __init__(self):
        foo_parser = argparse.ArgumentParser()

        foo_parser.add_argument('method')
        self.args = foo_parser.parse_args()

    def __call__(self, *args, **kws):
        method = self.args.method
        return getattr(self, method)(*args, **kws)

if __name__ == "__main__":
    main = Main()

    main()

example of usage:

$ python test.py foo
'foo'
$ python test.py bar
'bar'

回复收藏 0 原文

羞稚 2024-10-18 01:15:51

如果“方法”指的是“实例方法”：您无法使用实例调用实例方法，因此您首先需要一个实例。然后，您只需引用 obj.method 并获取一个 BoundMethod 来记住实例 (obj) - 例如：

class Cls:
    def __init__(self, x):
        self.x = x
    def foo(self, new_x):
        self.x = new_x

obj = Cls(1)
foo = obj.foo
foo(2) # same as obj.foo(), but we don't need to keep obj for this!
print(obj.x) #=>2

如果它是静态的 (如果是这样，为什么？99％的自由函数使用起来更简单并且工作得同样好）或类方法，只需参考Cls.method。

If by "method" you mean "instance method": You can't call an instance method with instance, so you need an instance first. Then, you can just refer to obj.method and get a BoundMethod which remembers the instance (obj) - e.g.:

class Cls:
    def __init__(self, x):
        self.x = x
    def foo(self, new_x):
        self.x = new_x

obj = Cls(1)
foo = obj.foo
foo(2) # same as obj.foo(), but we don't need to keep obj for this!
print(obj.x) #=>2

If it's a static (if so, why? in 99% free functions are simpler to use and work just as well) or class method, just refer to Cls.method.

回复收藏 0 原文

~没有更多了~