What is the difference between i = i + 1 and i += 1 in a 'for' loop? [duplicate]

I found out a curious thing today and was wondering if somebody could shed some light into what the difference is here?

import numpy as np

A = np.arange(12).reshape(4,3)
for a in A:
    a = a + 1

B = np.arange(12).reshape(4,3)
for b in B:
    b += 1

After running each for loop, A has not changed, but B has had one added to each element. I actually use the B version to write to a initialized NumPy array within a for loop.

Solution 1:

The difference is that one modifies the data-structure itself (in-place operation) b += 1 while the other just reassigns the variable a = a + 1.

Just for completeness:

x += y is not always doing an in-place operation, there are (at least) three exceptions:

If x doesn't implement an __iadd__ method then the x += y statement is just a shorthand for x = x + y. This would be the case if x was something like an int.
If __iadd__ returns NotImplemented, Python falls back to x = x + y.
The __iadd__ method could theoretically be implemented to not work in place. It'd be really weird to do that, though.

As it happens your bs are numpy.ndarrays which implements __iadd__ and return itself so your second loop modifies the original array in-place.

You can read more on this in the Python documentation of "Emulating Numeric Types".

These [__i*__] 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). 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, as with the evaluation of x + y. In certain situations, augmented assignment can result in unexpected errors (see Why does a_tuple[i] += ["item"] raise an exception when the addition works?), but this behavior is in fact part of the data model.

Solution 2:

In the first example, you are reassigning the variable a, while in the second one you are modifying the data in-place, using the += operator.

See the section about 7.2.1. Augmented assignment statements :

An augmented assignment expression like x += 1 can be rewritten as x = x + 1 to achieve a similar, but not exactly equal effect. In the augmented version, x is only evaluated once. Also, when possible, the actual operation is performed in-place, meaning that rather than creating a new object and assigning that to the target, the old object is modified instead.

+= operator calls __iadd__. This function makes the change in-place, and only after its execution, the result is set back to the object you are "applying" the += on.

__add__ on the other hand takes the parameters and returns their sum (without modifying them).

Solution 3:

As already pointed out, b += 1 updates b in-place, while a = a + 1 computes a + 1 and then assigns the name a to the result (now a does not refer to a row of A anymore).

To understand the += operator properly though, we need also to understand the concept of mutable versus immutable objects. Consider what happens when we leave out the .reshape:

C = np.arange(12)
for c in C:
    c += 1
print(C)  # [ 0  1  2  3  4  5  6  7  8  9 10 11]

We see that C is not updated, meaning that c += 1 and c = c + 1 are equivalent. This is because now C is a 1D array (C.ndim == 1), and so when iterating over C, each integer element is pulled out and assigned to c.

Now in Python, integers are immutable, meaning that in-place updates are not allowed, effectively transforming c += 1 into c = c + 1, where c now refers to a new integer, not coupled to C in any way. When you loop over the reshaped arrays, whole rows (np.ndarray's) are assigned to b (and a) at a time, which are mutable objects, meaning that you are allowed to stick in new integers at will, which happens when you do a += 1.

It should be mentioned that though + and += are meant to be related as described above (and very much usually are), any type can implement them any way it wants by defining the __add__ and __iadd__ methods, respectively.

Solution 4:

The short form(a += 1) has the option to modify a in-place , instead of creating a new object representing the sum and rebinding it back to the same name(a = a + 1).So,The short form(a += 1) is much efficient as it doesn't necessarily need to make a copy of a unlike a = a + 1.

Also even if they are outputting the same result, notice they are different because they are separate operators: + and +=