Question

我无法理解x和y是如何相同的列表。我一直在尝试使用print语句和import code; code.interact(local=locals())进行调试，以便进入各个方面，但我无法弄清楚究竟发生了什么： - （

from collections import namedtuple, OrderedDict

coordinates_2d=["x","y"]

def virtual_container(virtual_container, objects_type):
    """Used to create a virtual object given a the type of container and what it holds.
    The object_type needs to only have normal values."""
    if issubclass(virtual_container, list):
        class my_virtual_container_class:
            """This singleton class represents the container"""
            def __init__(self):
                #Define the default values
                __vals__=OrderedDict([(key,list()) for key in objects_type])
                print(id(__vals__["x"]), id(__vals__["y"]))#ids are different: 12911896 12911968
                #Then functions to access them
                d={key: lambda self: self.__vals__[key] for key in objects_type}
                d["__vals__"]=__vals__
                #Construct a named tuple from this
                self.attr=type('attr_cl',(), d)()
                print(id(self.attr.x()), id(self.attr.y()))#ids are same: 32904544 32904544
            #TODO: Define the operators __del__, setitem, getitem. Also append
        return my_virtual_container_class()

#Nice method of handling coordinates
coordinates=virtual_container(list, coordinates_2d)
x=coordinates.attr.x()
y=coordinates.attr.y()
x.append(1)
y.append(2)
print(x, y)#Prints [1, 2] [1, 2]

Answer 1

问题在于这一行：

d={key: lambda self: self.__vals__[key] for key in objects_type}

lambda使用变量key的值，但是这个值在调用lambda时已经改变了 - 所以所有lambdas实际上都会使用相同的键值。

这可以通过一个小技巧解决：将密钥作为默认参数值传递给lambda：

... lambda self, key=key: self.__vals__[key] ...

这可以确保key的值绑定到创建lambda时的值。

Answer 2

我认为以下行应该是这样的（但遗憾的是我无法测试，因为我没有Python 3可用）：

# Then functions to access them
d = dict((key, lambda self: self.__vals__[key]) for key in objects_type)

为什么这些列表相同？

2 个答案: