Add to list vs. Increment

Question

Given that in Python:

element = element + [0]

should be equal to:

element += [0]

Why does one modify a list and the other does not? Here is a example:

>>> a = [[0, 0], [0,0]]
>>> for element in a:
...     element = element + [0]
... 
>>> a
[[0, 0], [0, 0]]

a is not modified. But if I increment:

>>> a = [[0, 0], [0,0]]
>>> for element in a:
...     element += [0]
...
>>> a
[[0, 0, 0], [0, 0, 0]]

a is modified.

Thanks, Frank

Answer 1

This is a fun side-effect of += operatior, which calls __iadd__ instead of __add__.

The statement x = x + y is equivalent to x = x.__add__(y), while x += y is equivalent to x = x.__iadd__(y).

This lets the list class optimize += by extending the existing (ex, x += y is roughly equivalent to x.extend(y)) list instead of creating an entirely new list (which is what + needs to do).

For example:

>>> a = [1, 2, 3]
>>> original_a = a
>>> b = [1, 2, 3]
>>> original_b = b
>>> a += [4]
>>> b = b + [4]
>>> a is original_a
True
>>> b is original_b
False

You can see that using += maintains the identity of the left hand side (ie, a new list isn't created) while using + does not maintain the identity (ie, a new list is created).

For more, see: http://docs.python.org/library/operator.html#operator.iadd and the paragraph directly above the documentation for operator.iadd.

Answer 2

In the first case, element = element + [0], you are creating a new list.

In the second case, element += [0], you are modifying an existing list.

Since the list of lists, a, contains pointers to the elements, only modifying the elements will actually change things. (That is, creating a new list does not change the pointers in a.)

This is seen more clearly if we take a simple example showing how lists work:

>>> a = [1, 2, 3]
>>> b = a
>>> a = [4, 5, 6]
>>> a
[4, 5, 6]
>>> b
[1, 2, 3]
>>> a = [1, 2, 3]
>>> b = a
>>> a += [4, 5, 6]
>>> b
[1, 2, 3, 4, 5, 6]
>>> a
[1, 2, 3, 4, 5, 6]

Assigning a variable to a list simply assigns a pointer.

Answer 3

Adding to what others said, there is a difference in what these statements do:

element = element + [0]

does

element = element.__add__([0])

while

element += [0]

does

element = element.__iadd__([0])

__iadd__(), in this case, is free to determine what to return: the original object with a modification or a new object.

In the case of a immutable object, it must return a different one (e.g., a = b = 8; a += 9 => a is not b.

But in the case of a mutable object, such as a list, it normally modifies this one:

a = b = []
a += [8]

=> a is b.

This different behaviour reflects in your for loop:

for element in a:
   element = element + [0]

=> name element gets rebound to a different object; original one remains untouched

for element in a:
   element += [0]

=> original object, which is as well contained in the outer list, a, gets modified. The fact that element is reassigned is irrelevant; it is not used.