在类对象中，当您仅更改列表的一个条目时如何自动更新属性？

Question

我已经看到了一个非常相似的问题的答案：

In class object, how to auto update attributes?

我把代码贴在这里：

class SomeClass(object):
    def __init__(self, n):
        self.list = range(0, n)

    @property
    def list(self):
        return self._list
    @list.setter
    def list(self, val):
        self._list = val
        self._listsquare = [x**2 for x in self._list ]

    @property
    def listsquare(self):
        return self._listsquare
    @listsquare.setter
    def listsquare(self, val):
        self.list = [int(pow(x, 0.5)) for x in val]

>>> c = SomeClass(5)
>>> c.listsquare
[0, 1, 4, 9, 16]
>>> c.list
[0, 1, 2, 3, 4]
>>> c.list = range(0,6)
>>> c.list
[0, 1, 2, 3, 4, 5]
>>> c.listsquare
[0, 1, 4, 9, 16, 25]
>>> c.listsquare = [x**2 for x in range(0,10)]
>>> c.list
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

在这段代码中，当我使用以下方法更新列表时：

>>> c.list = [1, 2, 3, 4]

c.listsquare 将相应更新：

>>> c.listsquare
[1, 4, 9, 16]

但是当我尝试时：

>>> c.list[0] = 5
>>> c.list
[5, 2, 3, 4]

Listsquares 未更新：

>>> c.listsquare
[1, 4, 9, 16]

当我只更改列表中的一项时，如何使 listsquare 自动更新？

Answer 1

你可以做到这一点的一种方法是拥有一个私有助手_List 类，它几乎与内置的list 类完全一样，但也有一个owner 属性。每次为 _List 实例的元素之一赋值时，它就可以修改其 owner 的 listsquare 属性以使其保持最新。由于它仅供SomeClass 使用，因此可以嵌套在其中以提供更多封装。

class SomeClass(object):
    class _List(list):
        def __init__(self, owner, *args, **kwargs):
            super(SomeClass._List, self).__init__(*args, **kwargs)
            self.owner = owner

        def __setitem__(self, index, value):
            super(SomeClass._List, self).__setitem__(index, value)
            self.owner.listsquare[index] = int(value**2)

    def __init__(self, n):
        self.list = SomeClass._List(self, range(0, n))

    @property
    def list(self):
        return self._list
    @list.setter
    def list(self, val):
        self._list = val
        self._listsquare = [x**2 for x in self._list ]

    @property
    def listsquare(self):
        return self._listsquare
    @listsquare.setter
    def listsquare(self, val):
        self.list = [int(pow(x, 0.5)) for x in val]

c = SomeClass(5)
print(c.list)        # --> [0, 1, 2, 3, 4]
print(c.listsquare)  # --> [0, 1, 4, 9, 16]
c.list[0] = 5
print(c.list)        # --> [5, 1, 2, 3, 4]
print(c.listsquare)  # --> [25, 1, 4, 9, 16]

Answer 2

首先，我建议您不要通过访问单个 list.setter 来更改两个不同的属性。这样做的原因：

>>> c.list[0] = 5
>>> c.list
[5, 2, 3, 4]
>>> c.listsquare
[1, 4, 9, 16]

不行，你正在访问 c.list.__setitem__ 方法。

c.list[0] = 5 
is equal to
c.list.__setitem__(0, 5)
or
list.__setitem__(c.list, 0, 5)
and as such, the list.__setitem__ method isn't the one you've implemented in your class.

但是，如果你真的想这样做，你应该重新考虑基于 self._list 创建方形列表。

class SomeClass(object):
    def __init__(self, n):
        self.list = range(0, n)

    @property
    def list(self):
        return self._list

    @list.setter
    def list(self, val):
        self._list = val

    @property
    def listsquare(self):
        return [n ** 2 for n in self.list]

    @listsquare.setter
    def listsquare(self, val):
        self.list = [int(pow(x, 0.5)) for x in val]