Python operator.methodcaller（）无法正确处理参数？

我想在python函数内部使用operator.methodcaller（）函数。但是，传递参数列表时，我收到奇怪的错误消息。

python文档使我相信，这通常应该不是问题：

返回一个可调用对象，该对象在其操作数上调用方法名称。如果给出了其他参数和/或关键字参数，它们也将被赋予方法。例如：

在f = methodcaller（'name'）之后，调用f（b）返回b.name（）。

在f = methodcaller（'name'，'foo'，bar = 1）之后，调用f（b）返回b.name（'foo'，bar = 1）。

我在哪里错了？有没有其他简单的选择？

我编写了一个简短的代码示例来说明我的问题：

import operator

class CallTest:
    def __init__(self):
        self.x = 1
        self.y = 2

    def mutate(self, new_x, some_y = 2):
        self.x = new_x
        self.y = some_y

def map_something(func, iterable):
    """ This doesn't work and that's as expected. The error shows that the methodcaller is referencing
        the correct object though.
    """
    for i in iterable:
        func(i)

def map_something_else(func, iterable, \*rgs, \*kwargs):
    """ Why doesn't this work? And are there alternatives? """
    for i in iterable:
        func(i, \*rgs, \*kwargs)

def map_differently(func, iterable, \*rgs, \*kwargs):
    """ This works as expected but is kind of pointless in this case because i could replace this
        with getattr()
    """
    for i in iterable:
        operator.methodcaller(func, \*rgs, \* kwargs)(i)


if __name__ == '__main__':
    caller = operator.methodcaller('mutate')

    objs = [CallTest() for i in range(0, 2)]
    # map_something(caller, objs) # As expected we get: TypeError: mutate() missing 1 required positional argument: 'new_x'

    # map_something_else(caller, objs, 5) # TypeError: methodcaller expected 1 argument, got 2

    # map_something_else(caller, objs, 5, some_y=3) # TypeError: methodcaller() takes no keyword arguments

    map_differently('mutate', objs, 5, some_y=3) # This works as expected.

    for obj in objs:
        print(obj.x, obj.y)

问题来源：stackoverflow