使用 frozenset 的增强赋值

Question

我刚刚在冻结集上尝试了增广赋值，结果让我感到惊讶：

>>> x = frozenset(['foo', 'bar', 'baz'])
>>> x
frozenset({'foo', 'baz', 'bar'})
>>> x &= {'baz', 'qux', 'quux'}
>>> x
frozenset({'baz'})

这不应该发生，是吗？ frozensets 不是不可变的吗？

Answer 1

Frozensets 是不可变的，除了你的赋值没有改变原始的 frozenset - 你只是将变量 x 重新分配给你的二元运算符 &。正如 user2357112 在评论中指出的那样，在找不到 __iand__ 方法后，x &= {'baz', 'qux', 'quux'} 退回到 x = x & {'baz', 'qux', 'quux'}，留下 non-mutating 操作。

对于不提供 __iand__ 的不可变类型的其他增强操作，可以看到此行为，例如

In[1]: x = (1, 2, 3)
In[2]: id(x)
Out[2]: 761782862328
In[3]: x += (4, 5)
In[4]: id(x)   # we now have a different id
Out[4]: 761780182888
In[5]: x[2] = 3  # an actual mutating operation
TypeError: 'tuple' object does not support item assignment

Answer 2

你为什么感到惊讶？

您知道 "augmented assignment" 这个词，所以找到 "Python Data Model on augmented arithmetic assignments"（强调我的）没有问题：

These [__i***__] methods should attempt to do the operation in-place (modifying self) and return the result (which could be, but does not have to be, self). If a specific method is not defined, the augmented assignment falls back to the normal methods. For instance, if x is an instance of a class with an __iadd__() method, x += y is equivalent to x = x.__iadd__(y) . Otherwise, x.__add__(y) and y.__radd__(x) are considered, [...]

>>> x = frozenset(['foo', 'bar', 'baz'])
>>> x.__iand__
[...]
AttributeError: 'frozenset' object has no attribute '__iand__'

所以它没有 __iand__ 方法所以你执行的代码是：

>>> x = x & {'baz', 'qux', 'quux'}

然而 __and__ 方法由 frozenset:

定义

>>> x & {'baz', 'qux', 'quux'}
frozenset({'baz'})

但是您丢失了对原始 frozenset 的引用：x：

>>> y = x   # that doesn't do a copy, it's just to check if `x` has changed"
>>> x &= {'baz', 'qux', 'quux'}
>>> x is y  # check if they reference the same object!
False
>>> x, y
(frozenset({'baz'}), frozenset({'bar', 'baz', 'foo'}))

但这只是在 "Principle of least astonishment" 之后。您想要 __and__ 并且明确表示您不想保留原来的 x - in-place 操作也会改变它！

再说一遍：为什么这让您感到惊讶？

使用 frozenset 的增强赋值

Augmented assignment with frozenset

python

set

immutability

augmented-assignment