Augmented assignment with frozenset

Question

I just tried an augmented assignment on a frozenset, and the result surprised me:

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

This isn't supposed to happen, is it? Aren't frozensets immutable?

Answer 1

Frozensets are immutable, except your assignment isn't mutating the original frozenset - you are just reassigning the variable x to the result of your binary operator &. As noted by user2357112 in the comments, x &= {'baz', 'qux', 'quux'} falls back on x = x & {'baz', 'qux', 'quux'} after an __iand__ method is not found, leaving you with a non-mutating operation.

This behavior can be seen for other augmented operations on immutable types that don't supply __iand__, e.g.

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

Why are you surprised?

You knew the term "augmented assignment" so there is no problem finding the "Python Data Model on augmented arithmetic assignments" (emphasis mine):

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__'

So it has no __iand__ method so the code you perform is:

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

The __and__ method however is defined by frozenset:

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

However you lost your reference to the original frozenset: 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'}))

But that just follows the "Principle of least astonishment". You wanted the __and__ and you made it clear that you didn't want to keep your original x - an in-place operation also would have altered it!

So again: Why did that surprise you?