如何在Python中乘以函数? [英] How to multiply functions in python?

查看:171
本文介绍了如何在Python中乘以函数?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

  def sub3(n):
return n - 3

def square(n):
return n * n

在python中编写函数非常简单:

 >>> my_list 
[0,1,2,3,4,5,6,7,8,9]
>>> [my_list中的n的[square(sub3(n))]
[9,4,1,0,1,4,9,16,25,36]

不幸的是,当想将组合作为使用时,这有点蹩脚:

 >>> sort(my_list,key = lambda n:square(sub3(n)))
[3,2,4,1,5,0,6,7,8,9]

这实际上应该是 sorted(my_list,key = square * sub3),因为heck,函数 __ mul __ 不会用于其他任何事情:

 >>> square * sub3 
TypeError:*:'function'和'function'不支持的操作数类型


$ b $

 >>>好的,让我们来定义它吧! type(sub3).__ mul__ ='something'
TypeError:无法设置内置/扩展类型的属性'function'

D'oh!

 >>> class CoolerFunction(types.FunctionType):
...传递
...
TypeError:调用元类时出错
类型'function'不是可接受的基类型

D'oh!

  class Hack(object):
def __init __(self,function):
self.function = function
def __call __(self,* args,** kwargs):
返回self.function(* args,** kwargs)
def __mul __(self,other):
def hack(* args,** kwargs):
返回self。函数(其他(* args,** kwargs))
return Hack(黑客)

嘿,现在我们到了某个地方..

 >>> square = Hack(square)
>>> sub3 = Hack(sub3)
>>> [my_list中的n的[square(sub3(n))]
[9,4,1,0,1,4,9,16,25,36]
>>> [my_list中的n的[(square * sub3)(n)]
[9,4,1,0,1,4,9,16,25,36]
>>>排序(my_list,key = square * sub3)
[3,2,4,1,5,0,6,7,8,9]

但我不想要一个 Hack callable class!我没有完全理解的范围规则是完全不同的,甚至比lameda更丑陋。我想monkeypatch 功能。我可以怎么做?

解决方案

你可以使用你的hack类作为装饰器,可能想为班级选择一个更合适的名称。



像这样:

  b $ self_function = function $ b $ def __call __(self,* args,** kwargs):
返回self.function(* args,** kwargs)
def __mul __(self,other):
@Composable
def构成(* args,** kwargs):
返回self.function(其他(* args,** kwargs))
返回组合
def __rmul __(self,other):
@Composable
def组合(* args, ** kwargs):
return other(self.function(* args,** kwargs))
返回组合

然后你可以像这样装饰你的函数:

  @Composable 
def sub3 (n):
return n - 3

@Composable
def square(n):
return n * n


$ b

 

square * sub3)(n)

基本上你用hack类完成了同样的事情,但将其用作装饰器。


def sub3(n):
    return n - 3

def square(n):
    return n * n

It's dead easy to compose functions in python:

>>> my_list
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> [square(sub3(n)) for n in my_list]
[9, 4, 1, 0, 1, 4, 9, 16, 25, 36]

Unfortunately, when wanting to use the composition as a key, it's kind of lame:

>>> sorted(my_list, key=lambda n: square(sub3(n)))
[3, 2, 4, 1, 5, 0, 6, 7, 8, 9]

This should really just be sorted(my_list, key=square*sub3), because heck, function __mul__ isn't used for anything else anyway:

>>> square * sub3
TypeError: unsupported operand type(s) for *: 'function' and 'function'

Well let's just define it then!

>>> type(sub3).__mul__ = 'something'
TypeError: can't set attributes of built-in/extension type 'function'

D'oh!

>>> class CoolerFunction(types.FunctionType):
...     pass
...
TypeError: Error when calling the metaclass bases
    type 'function' is not an acceptable base type

D'oh!

class Hack(object):
    def __init__(self, function):
        self.function = function
    def __call__(self, *args, **kwargs):
        return self.function(*args, **kwargs)
    def __mul__(self, other):
        def hack(*args, **kwargs):
            return self.function(other(*args, **kwargs))
        return Hack(hack)

Hey, now we're getting somewhere..

>>> square = Hack(square)
>>> sub3 = Hack(sub3)
>>> [square(sub3(n)) for n in my_list]
[9, 4, 1, 0, 1, 4, 9, 16, 25, 36]
>>> [(square*sub3)(n) for n in my_list]
[9, 4, 1, 0, 1, 4, 9, 16, 25, 36]
>>> sorted(my_list, key=square*sub3)
[3, 2, 4, 1, 5, 0, 6, 7, 8, 9]

But I don't want a Hack callable class! The scoping rules are totally different in ways I don't fully understand, and this is even uglier than the "lameda" arguably. I want to monkeypatch the functions. How can I do that?

解决方案

You can use your hack class as a decorator pretty much as it's written, though you'd likely want to choose a more appropriate name for the class.

Like this:

class Composable(object):
    def __init__(self, function):
        self.function = function
    def __call__(self, *args, **kwargs):
        return self.function(*args, **kwargs)
    def __mul__(self, other):
        @Composable
        def composed(*args, **kwargs):
            return self.function(other(*args, **kwargs))
        return composed
    def __rmul__(self, other):
        @Composable
        def composed(*args, **kwargs):
            return other(self.function(*args, **kwargs))
        return composed

You can then decorate your functions like so:

@Composable
def sub3(n):
    return n - 3

@Composable
def square(n):
    return n * n

And compose them like so:

(square * sub3)(n)

Basically it's the same thing you've accomplished using your hack class, but using it as a decorator.

这篇关于如何在Python中乘以函数?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆