这实际上是对“杰森·普拉特”答案的补充

虽然Jasons的答案是有效的，但它只在想要向类中添加函数时有效。当我试图从.py源代码文件重新加载一个已经存在的方法时，它对我来说并不起作用。

我花了很长时间才找到解决方法，但技巧似乎很简单。。。1.从源代码文件导入代码2.强制重新加载3.rd使用types.FunctionType（…）将导入和绑定的方法转换为函数您还可以传递当前的全局变量，因为重新加载的方法将位于不同的命名空间中4.现在你可以按照“杰森·普拉特”的建议继续使用类型.MethodType（…）

例子：

# this class resides inside ReloadCodeDemo.py
class A:
    def bar( self ):
        print "bar1"
        
    def reloadCode(self, methodName):
        ''' use this function to reload any function of class A'''
        import types
        import ReloadCodeDemo as ReloadMod # import the code as module
        reload (ReloadMod) # force a reload of the module
        myM = getattr(ReloadMod.A,methodName) #get reloaded Method
        myTempFunc = types.FunctionType(# convert the method to a simple function
                                myM.im_func.func_code, #the methods code
                                globals(), # globals to use
                                argdefs=myM.im_func.func_defaults # default values for variables if any
                                ) 
        myNewM = types.MethodType(myTempFunc,self,self.__class__) #convert the function to a method
        setattr(self,methodName,myNewM) # add the method to the function

if __name__ == '__main__':
    a = A()
    a.bar()
    # now change your code and save the file
    a.reloadCode('bar') # reloads the file
    a.bar() # now executes the reloaded code

自python 2.6以来，模块new已弃用，并在3.0中删除，请使用类型

看见http://docs.python.org/library/new.html

在下面的示例中，我故意从patch_me（）函数中删除了返回值。我认为，给出返回值可能会让人相信补丁会返回一个新对象，这是不正确的——它会修改传入的对象。这可能有助于更严格地使用猴痘。

import types

class A(object):#but seems to work for old style objects too
    pass

def patch_me(target):
    def method(target,x):
        print "x=",x
        print "called from", target
    target.method = types.MethodType(method,target)
    #add more if needed

a = A()
print a
#out: <__main__.A object at 0x2b73ac88bfd0>  
patch_me(a)    #patch instance
a.method(5)
#out: x= 5
#out: called from <__main__.A object at 0x2b73ac88bfd0>
patch_me(A)
A.method(6)        #can patch class too
#out: x= 6
#out: called from <class '__main__.A'>

整合Jason Pratt和社区wiki的答案，看看不同绑定方法的结果：

特别注意将绑定函数添加为类方法是如何工作的，但引用范围不正确。

#!/usr/bin/python -u
import types
import inspect

## dynamically adding methods to a unique instance of a class


# get a list of a class's method type attributes
def listattr(c):
    for m in [(n, v) for n, v in inspect.getmembers(c, inspect.ismethod) if isinstance(v,types.MethodType)]:
        print m[0], m[1]

# externally bind a function as a method of an instance of a class
def ADDMETHOD(c, method, name):
    c.__dict__[name] = types.MethodType(method, c)

class C():
    r = 10 # class attribute variable to test bound scope

    def __init__(self):
        pass

    #internally bind a function as a method of self's class -- note that this one has issues!
    def addmethod(self, method, name):
        self.__dict__[name] = types.MethodType( method, self.__class__ )

    # predfined function to compare with
    def f0(self, x):
        print 'f0\tx = %d\tr = %d' % ( x, self.r)

a = C() # created before modified instnace
b = C() # modified instnace


def f1(self, x): # bind internally
    print 'f1\tx = %d\tr = %d' % ( x, self.r )
def f2( self, x): # add to class instance's .__dict__ as method type
    print 'f2\tx = %d\tr = %d' % ( x, self.r )
def f3( self, x): # assign to class as method type
    print 'f3\tx = %d\tr = %d' % ( x, self.r )
def f4( self, x): # add to class instance's .__dict__ using a general function
    print 'f4\tx = %d\tr = %d' % ( x, self.r )


b.addmethod(f1, 'f1')
b.__dict__['f2'] = types.MethodType( f2, b)
b.f3 = types.MethodType( f3, b)
ADDMETHOD(b, f4, 'f4')


b.f0(0) # OUT: f0   x = 0   r = 10
b.f1(1) # OUT: f1   x = 1   r = 10
b.f2(2) # OUT: f2   x = 2   r = 10
b.f3(3) # OUT: f3   x = 3   r = 10
b.f4(4) # OUT: f4   x = 4   r = 10


k = 2
print 'changing b.r from {0} to {1}'.format(b.r, k)
b.r = k
print 'new b.r = {0}'.format(b.r)

b.f0(0) # OUT: f0   x = 0   r = 2
b.f1(1) # OUT: f1   x = 1   r = 10  !!!!!!!!!
b.f2(2) # OUT: f2   x = 2   r = 2
b.f3(3) # OUT: f3   x = 3   r = 2
b.f4(4) # OUT: f4   x = 4   r = 2

c = C() # created after modifying instance

# let's have a look at each instance's method type attributes
print '\nattributes of a:'
listattr(a)
# OUT:
# attributes of a:
# __init__ <bound method C.__init__ of <__main__.C instance at 0x000000000230FD88>>
# addmethod <bound method C.addmethod of <__main__.C instance at 0x000000000230FD88>>
# f0 <bound method C.f0 of <__main__.C instance at 0x000000000230FD88>>

print '\nattributes of b:'
listattr(b)
# OUT:
# attributes of b:
# __init__ <bound method C.__init__ of <__main__.C instance at 0x000000000230FE08>>
# addmethod <bound method C.addmethod of <__main__.C instance at 0x000000000230FE08>>
# f0 <bound method C.f0 of <__main__.C instance at 0x000000000230FE08>>
# f1 <bound method ?.f1 of <class __main__.C at 0x000000000237AB28>>
# f2 <bound method ?.f2 of <__main__.C instance at 0x000000000230FE08>>
# f3 <bound method ?.f3 of <__main__.C instance at 0x000000000230FE08>>
# f4 <bound method ?.f4 of <__main__.C instance at 0x000000000230FE08>>

print '\nattributes of c:'
listattr(c)
# OUT:
# attributes of c:
# __init__ <bound method C.__init__ of <__main__.C instance at 0x0000000002313108>>
# addmethod <bound method C.addmethod of <__main__.C instance at 0x0000000002313108>>
# f0 <bound method C.f0 of <__main__.C instance at 0x0000000002313108>>

就我个人而言，我更喜欢外部ADDMETHOD函数路由，因为它也允许我在迭代器中动态分配新的方法名。

def y(self, x):
    pass
d = C()
for i in range(1,5):
    ADDMETHOD(d, y, 'f%d' % i)
print '\nattributes of d:'
listattr(d)
# OUT:
# attributes of d:
# __init__ <bound method C.__init__ of <__main__.C instance at 0x0000000002303508>>
# addmethod <bound method C.addmethod of <__main__.C instance at 0x0000000002303508>>
# f0 <bound method C.f0 of <__main__.C instance at 0x0000000002303508>>
# f1 <bound method ?.y of <__main__.C instance at 0x0000000002303508>>
# f2 <bound method ?.y of <__main__.C instance at 0x0000000002303508>>
# f3 <bound method ?.y of <__main__.C instance at 0x0000000002303508>>
# f4 <bound method ?.y of <__main__.C instance at 0x0000000002303508>>

我认为上述答案没有抓住重点。

让我们用一个方法来上课：

class A(object):
    def m(self):
        pass

现在，让我们在ipython中玩一下：

In [2]: A.m
Out[2]: <unbound method A.m>

好的，所以m（）在某种程度上变成了A的非绑定方法，但它真的是这样吗？

In [5]: A.__dict__['m']
Out[5]: <function m at 0xa66b8b4>

事实证明，m（）只是一个函数，对它的引用被添加到a类字典中——这没有什么魔力。那为什么A.m会给我们一个未绑定的方法？这是因为点没有被翻译成简单的字典查找。这实际上是对a.__class__.__getattribute__（a，'m'）的调用：

In [11]: class MetaA(type):
   ....:     def __getattribute__(self, attr_name):
   ....:         print str(self), '-', attr_name

In [12]: class A(object):
   ....:     __metaclass__ = MetaA

In [23]: A.m
<class '__main__.A'> - m
<class '__main__.A'> - m

现在，我不清楚为什么最后一行要打印两次，但仍然很清楚是怎么回事。

现在，默认__getattribute__所做的是检查属性是否是所谓的描述符，即它是否实现了一个特殊的__get__方法。如果它实现了该方法，那么返回的是调用__get__方法的结果。回到我们A类的第一个版本，这是我们拥有的：

In [28]: A.__dict__['m'].__get__(None, A)
Out[28]: <unbound method A.m>

因为Python函数实现了描述符协议，所以如果代表对象调用它们，它们会在__get__方法中将自己绑定到该对象。

好的，那么如何向现有对象添加方法呢？假设您不介意修补类，那么简单如下：

B.m = m

然后，由于描述符的魔力，B.m“成为”一个未绑定的方法。

如果你想将一个方法添加到一个对象中，那么你必须自己使用types.MethodType来模拟机器：

b.m = types.MethodType(m, b)

顺便说一句：

In [2]: A.m
Out[2]: <unbound method A.m>

In [59]: type(A.m)
Out[59]: <type 'instancemethod'>

In [60]: type(b.m)
Out[60]: <type 'instancemethod'>

In [61]: types.MethodType
Out[61]: <type 'instancemethod'>

除了其他人所说的，我发现__repr_和__str__方法不能在对象级别上进行猴痘，因为repr（）和str（）使用类方法，而不是本地绑定的对象方法：

# Instance monkeypatch
[ins] In [55]: x.__str__ = show.__get__(x)                                                                 

[ins] In [56]: x                                                                                           
Out[56]: <__main__.X at 0x7fc207180c10>

[ins] In [57]: str(x)                                                                                      
Out[57]: '<__main__.X object at 0x7fc207180c10>'

[ins] In [58]: x.__str__()                                                                                 
Nice object!

# Class monkeypatch
[ins] In [62]: X.__str__ = lambda _: "From class"                                                          

[ins] In [63]: str(x)                                                                                      
Out[63]: 'From class'

由于这个问题是针对非Python版本提出的，这里是JavaScript：

a.methodname = function () { console.log("Yay, a new method!") }