将外部函数分配给Python中的类变量 [英] Assign external function to class variable in Python

问题描述

我试图将在其他地方定义的函数分配给类变量，以便以后可以在实例的方法之一中调用它，如下所示:

I am trying to assign a function defined elsewhere to a class variable so I can later call it in one of the methods of the instance, like this:

from module import my_func

class Bar(object):
    func = my_func
    def run(self):
        self.func()  # Runs my function

问题在于此操作失败，因为执行self.func()时，实例将作为第一个参数传递.

The problem is that this fails because when doing self.func(), then the instance is passed as the first parameter.

我想出了一个办法，但是对我来说似乎很丑陋，有人可以替代吗?

I've come up with a hack but seems ugly to me, anybody has an alternative?

In [1]: class Foo(object):
   ...:     func = lambda *args: args
   ...:     def __init__(self):
   ...:         print(self.func())
   ...:

In [2]: class Foo2(object):
   ...:     funcs = [lambda *args: args]
   ...:     def __init__(self):
   ...:         print(self.funcs[0]())
   ...:

In [3]: f = Foo()
(<__main__.Foo object at 0x00000000044BFB70>,)

In [4]: f2 = Foo2()
()

编辑:行为与内置函数不同！

The behavior is different with builtin functions!

In [13]: from math import pow

In [14]: def pow_(a, b):
   ....:     return pow(a, b)
   ....:

In [15]: class Foo3(object):
   ....:     func = pow_
   ....:     def __init__(self):
   ....:         print(self.func(2, 3))
   ....:

In [16]: f3 = Foo3()
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-16-c27c8778655e> in <module>()
----> 1 f3 = Foo3()

<ipython-input-15-efeb6adb211c> in __init__(self)
      2     func = pow_
      3     def __init__(self):
----> 4         print(self.func(2, 3))
      5

TypeError: pow_() takes exactly 2 arguments (3 given)

In [17]: class Foo4(object):
   ....:     func = pow
   ....:     def __init__(self):
   ....:         print(self.func(2, 3))
   ....:

In [18]: f4 = Foo4()
8.0

推荐答案

Python函数是 descriptor 对象，当类中的属性访问它们时，实例将它们绑定为方法.

Python functions are descriptor objects, and when attributes on a class accessing them an instance causes them to be bound as methods.

如果要防止这种情况，请使用 staticmethod函数将函数包装在不绑定到实例的其他描述符中:

If you want to prevent this, use the staticmethod function to wrap the function in a different descriptor that doesn't bind to the instance:

class Bar(object):
    func = staticmethod(my_func)
    def run(self):
        self.func()

或者，通过方法的__func__属性访问未绑定函数:

Alternatively, access the unbound function via the __func__ attribute on the method:

def run(self):
    self.func.__func__()

或直接转到class __dict__属性以完全绕过描述符协议:

or go directly to the class __dict__ attribute to bypass the descriptor protocol altogether:

def run(self):
    Bar.__dict__['func']()

对于math.pow来说，它不是 Python 函数，因为它是用C代码编写的.大多数内置函数都是用C编写的，大多数不是描述符.

As for math.pow, that's not a Python function, in that it is written in C code. Most built-in functions are written in C, and most are not descriptors.

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