Python a &= b meaning?

Question

What does the &= operator mean in Python, and can you give me a working example?

I am trying to understand the __iand__ operator.

I just don't know what &= means and have looked online but couldn't find it.

Answer 1

Explanation

Understandable that you can't find much reference on it. I find it hard to get references on this too, but they exist.

The i in iand means in-place, so it's the in-place operator for &. &= calls the __iand__ operator, if it is implemented. If not implemented, it is the same as x = x & y.

Built-in Example, Sets:

It's primarily used to update the intersection of built-in set types:

>>> a = set('abc') 
>>> a &= set('cbe')
>>> a
set(['c', 'b'])

which is the same as:

>>> a = set('abc')
>>> a.__iand__(set('cbe'))
set(['c', 'b'])

It is very similar to calling the set.intersection_update method, and would be used in control flow as you would do an in-place update of any object or variable (if the object is immutable).

Unimplemented Built-in Example

The less commonly used immutable frozenset object would be replaced in memory on the inplace update, and the variable name would point to the new object in memory.

>>> a = frozenset('abc')
>>> a &= set('bce')
>>> a
frozenset({'c', 'b'})

In this case, since frozenset doesn't implement an __iand__ method,

>>> a = frozenset('abc')
>>> a.__iand__(set('cbe'))

Traceback (most recent call last):
  File "<pyshell#160>", line 1, in <module>
    a = frozenset('abc'); a.__iand__(set('cbe'))
AttributeError: 'frozenset' object has no attribute '__iand__'

it is (*nearly) identical to

a = a & set('bce')

*(I say nearly because if you examine the bytecode, you'll see that the underlying implementation treats sets and frozensets the same, even though frozensets don't have __iand__, and sets do, because each calls INPLACE_AND, at least for compiled functions.)

Built-in Example, Binary Flags:

Similar to Sets, we can use the &= to update the intersection of binary option flags where the value for True is 1. Below, we demonstrate that the "binary AND", (akin to intersection) of the binary numbers 1110 and 1011 is 1010:

>>> option_flags = int('1110', 2)
>>> option_flags
14
>>> option_flags &= int('1011', 2)
>>> option_flags
10
>>> bin(option_flags)
'0b1010'

Since int objects are not mutable, like the frozenset example, this actually only reassigns the variable option_flags to the newly calculated value.

Answer 2

Contrary to some of the other answers, a &= b is not shorthand for a = a & b, though I admit it often behaves similarly for built-in immutable types like integers.

a &= b can call the special method __iand__ if available. To see the difference, let's define a custom class:

class NewIand(object):
    def __init__(self, x):
        self.x = x
    def __and__(self, other):
        return self.x & other.x
    def __iand__(self, other):
        return 99  

After which we have

>>> a = NewIand(1+2+4)
>>> b = NewIand(4+8+16)
>>> a & b
4
>>> a = a & b
>>> a
4

but

>>> a = NewIand(1+2+4)
>>> b = NewIand(4+8+16)
>>> a &= b
>>> a
99