Question

Consider the following session. How are the differences explained? I thought that a += b is a syntactical sugar of (and thus equivalent to) a = a + b. Obviously I’m wrong.

>>> import numpy as np
>>> a  = np.arange(24.).reshape(4,6)
>>> print a
[[  0.   1.   2.   3.   4.   5.]
 [  6.   7.   8.   9.  10.  11.]
 [ 12.  13.  14.  15.  16.  17.]
 [ 18.  19.  20.  21.  22.  23.]]
>>> for line in a:
...     line += 100
...
>>> print a #a has been changed
[[ 100.  101.  102.  103.  104.  105.]
 [ 106.  107.  108.  109.  110.  111.]
 [ 112.  113.  114.  115.  116.  117.]
 [ 118.  119.  120.  121.  122.  123.]]
>>>
>>> for line in a:
...     line = line + 999
...
>>> print a #a hasn't been changed
[[ 100.  101.  102.  103.  104.  105.]
 [ 106.  107.  108.  109.  110.  111.]
 [ 112.  113.  114.  115.  116.  117.]
 [ 118.  119.  120.  121.  122.  123.]]

Thank you

Answer 1

Using the + operator results in a call to the special method __add__ which should create a new object and should not modify the original.

On the other hand, using the += operator results in a call to __iadd__ which should modify the object if possible rather than creating a new object.

__add__

These methods are called to implement the binary arithmetic operations (+, -, *, //, %, divmod(), pow(), **, <<, >>, &, ^, |). For instance, to evaluate the expression x + y, where x is an instance of a class that has an __add__() method, x.__add__(y) is called.

__iadd__

These methods are called to implement the augmented arithmetic assignments (+=, -=, *=, /=, //=, %=, **=, <<=, >>=, &=, ^=, |=). These 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).

Of course it is possible to implement __add__ and __iadd__ to have some other behaviour if you wanted to, but what you observe is the standard and recommended way. And, yes, it is a little surprising the first time you see it.