标签归档:bind

Python:绑定未绑定方法?

问题:Python:绑定未绑定方法?

在Python中,有没有办法绑定未绑定的方法而不调用它?

我正在编写一个wxPython程序,对于某个类,我决定将所有按钮的数据分组为一个类级别的元组列表是很不错的,如下所示:

class MyWidget(wx.Window):
    buttons = [("OK", OnOK),
               ("Cancel", OnCancel)]

    # ...

    def Setup(self):
        for text, handler in MyWidget.buttons:

            # This following line is the problem line.
            b = wx.Button(parent, label=text).Bind(wx.EVT_BUTTON, handler)

问题是,因为所有的值handler都是未绑定方法,所以我的程序爆炸得很厉害,我哭了。

我正在网上寻找解决方案,该方案应该是一个相对简单,可解决的问题。不幸的是我找不到任何东西。现在,我正在functools.partial解决此问题,但是没有人知道是否存在一种干净,健康,Pythonic的方式将未绑定的方法绑定到实例并继续传递它而不调用它吗?

In Python, is there a way to bind an unbound method without calling it?

I am writing a wxPython program, and for a certain class I decided it’d be nice to group the data of all of my buttons together as a class-level list of tuples, like so:

class MyWidget(wx.Window):
    buttons = [("OK", OnOK),
               ("Cancel", OnCancel)]

    # ...

    def Setup(self):
        for text, handler in MyWidget.buttons:

            # This following line is the problem line.
            b = wx.Button(parent, label=text).Bind(wx.EVT_BUTTON, handler)

The problem is, since all of the values of handler are unbound methods, my program explodes in a spectacular blaze and I weep.

I was looking around online for a solution to what seems like should be a relatively straightforward, solvable problem. Unfortunately I couldn’t find anything. Right now, I’m using functools.partial to work around this, but does anyone know if there’s a clean-feeling, healthy, Pythonic way to bind an unbound method to an instance and continue passing it around without calling it?


回答 0

所有函数也是描述符,因此您可以通过调用它们的__get__方法来绑定它们:

bound_handler = handler.__get__(self, MyWidget)

这是R. Hettinger 关于描述符的出色指南


作为一个独立的例子,请参考Keith的 评论

def bind(instance, func, as_name=None):
    """
    Bind the function *func* to *instance*, with either provided name *as_name*
    or the existing name of *func*. The provided *func* should accept the 
    instance as the first argument, i.e. "self".
    """
    if as_name is None:
        as_name = func.__name__
    bound_method = func.__get__(instance, instance.__class__)
    setattr(instance, as_name, bound_method)
    return bound_method

class Thing:
    def __init__(self, val):
        self.val = val

something = Thing(21)

def double(self):
    return 2 * self.val

bind(something, double)
something.double()  # returns 42

All functions are also descriptors, so you can bind them by calling their __get__ method:

bound_handler = handler.__get__(self, MyWidget)

Here’s R. Hettinger’s excellent guide to descriptors.


As a self-contained example pulled from Keith’s comment:

def bind(instance, func, as_name=None):
    """
    Bind the function *func* to *instance*, with either provided name *as_name*
    or the existing name of *func*. The provided *func* should accept the 
    instance as the first argument, i.e. "self".
    """
    if as_name is None:
        as_name = func.__name__
    bound_method = func.__get__(instance, instance.__class__)
    setattr(instance, as_name, bound_method)
    return bound_method

class Thing:
    def __init__(self, val):
        self.val = val

something = Thing(21)

def double(self):
    return 2 * self.val

bind(something, double)
something.double()  # returns 42

回答 1

可以使用types.MethodType干净地完成此操作。例:

import types

def f(self): print self

class C(object): pass

meth = types.MethodType(f, C(), C) # Bind f to an instance of C
print meth # prints <bound method C.f of <__main__.C object at 0x01255E90>>

This can be done cleanly with types.MethodType. Example:

import types

def f(self): print self

class C(object): pass

meth = types.MethodType(f, C(), C) # Bind f to an instance of C
print meth # prints <bound method C.f of <__main__.C object at 0x01255E90>>

回答 2

创建一个带有self的闭包在技术上不会绑定该函数,但这是解决相同(或非常相似)潜在问题的另一种方法。这是一个简单的例子:

self.method = (lambda self: lambda args: self.do(args))(self)

Creating a closure with self in it will not technically bind the function, but it is an alternative way of solving the same (or very similar) underlying problem. Here’s a trivial example:

self.method = (lambda self: lambda args: self.do(args))(self)

回答 3

这将绑定selfhandler

bound_handler = lambda *args, **kwargs: handler(self, *args, **kwargs)

这是通过将self第一个参数传递给函数来实现的。object.function()只是语法糖function(object)

This will bind self to handler:

bound_handler = lambda *args, **kwargs: handler(self, *args, **kwargs)

This works by passing self as the first argument to the function. object.function() is just syntactic sugar for function(object).


回答 4

晚了,但是我来这里有一个类似的问题:我有一个类方法和一个实例,并且想要将该实例应用于该方法。

冒着过于简化OP问题的风险,我最终做了一些不太神秘的事情,这可能会对到这里来的其他人有用(注意:我正在使用Python 3 – YMMV工作)。

考虑这个简单的类:

class Foo(object):

    def __init__(self, value):
        self._value = value

    def value(self):
        return self._value

    def set_value(self, value):
        self._value = value

这是您可以使用的方法:

>>> meth = Foo.set_value   # the method
>>> a = Foo(12)            # a is an instance with value 12
>>> meth(a, 33)            # apply instance and method
>>> a.value()              # voila - the method was called
33

Late to the party, but I came here with a similar question: I have a class method and an instance, and want to apply the instance to the method.

At the risk of oversimplifying the OP’s question, I ended up doing something less mysterious that may be useful to others who arrive here (caveat: I’m working in Python 3 — YMMV).

Consider this simple class:

class Foo(object):

    def __init__(self, value):
        self._value = value

    def value(self):
        return self._value

    def set_value(self, value):
        self._value = value

Here’s what you can do with it:

>>> meth = Foo.set_value   # the method
>>> a = Foo(12)            # a is an instance with value 12
>>> meth(a, 33)            # apply instance and method
>>> a.value()              # voila - the method was called
33