Difference between a -= b and a = a - b in Python

Question

I have recently applied this solution for averaging every N rows of matrix. Although the solution works in general I had problems when applied to a 7x1 array. I have noticed tha

Answer 1

Note: using in-place operations on NumPy arrays that share memory in no longer a problem in version 1.13.0 onward (see details here). The two operation will produce the same result. This answer only applies to earlier versions of NumPy.

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Mutating arrays while they're being used in computations can lead to unexpected results!

In the example in the question, subtraction with -= modifies the second element of a and then immediately uses that modified second element in the operation on the third element of a.

Here is what happens with a[1:] -= a[:-1] step by step:

• a is the array with the data [1, 2, 3].

• We have two views onto this data: a[1:] is [2, 3], and a[:-1] is [1, 2].

• The in-place subtraction -= begins. The first element of a[:-1], 1, is subtracted from the first element of a[1:]. This has modified a to be [1, 1, 3]. Now we have that a[1:] is a view of the data [1, 3], and a[:-1] is a view of the data [1, 1] (the second element of array a has been changed).

• a[:-1] is now [1, 1] and NumPy must now subtract its second element which is 1 (not 2 anymore!) from the second element of a[1:]. This makes a[1:] a view of the values [1, 2].

• a is now an array with the values [1, 1, 2].

b[1:] = b[1:] - b[:-1] does not have this problem because b[1:] - b[:-1] creates a new array first and then assigns the values in this array to b[1:]. It does not modify b itself during the subtraction, so the views b[1:] and b[:-1] do not change.

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The general advice is to avoid modifying one view inplace with another if they overlap. This includes the operators -=, *=, etc. and using the out parameter in universal functions (like np.subtract and np.multiply) to write back to one of the arrays.

Answer 2

Internally, the difference is that this:

a[1:] -= a[:-1]

is equivalent to this:

a[1:] = a[1:].__isub__(a[:-1])
a.__setitem__(slice(1, None, None), a.__getitem__(slice(1, None, None)).__isub__(a.__getitem__(slice(1, None, None)))

while this:

b[1:] = b[1:] - b[:-1]

maps to this:

b[1:] = b[1:].__sub__(b[:-1])
b.__setitem__(slice(1, None, None), b.__getitem__(slice(1, None, None)).__sub__(b.__getitem__(slice(1, None, None)))

In some cases, __sub__() and __isub__() work in a similar way. But mutable objects should mutate and return themselves when using __isub__(), while they should return a new object with __sub__().

Applying slice operations on numpy objects creates views on them, so using them directly accesses the memory of the "original" object.

Answer 3

The docs say :

The idea behind augmented assignment in Python is that it isn't just an easier way to write the common practice of storing the result of a binary operation in its left-hand operand, but also a way for the left-hand operand in question to know that it should operate `on itself', rather than creating a modified copy of itself.

As a thumb rule, augmented substraction (x-=y) is x.__isub__(y), for IN-place operation IF possible, when normal substraction (x = x-y) is x=x.__sub__(y) . On non mutable objects like integers it's equivalent. But for mutable ones like arrays or lists, as in your example, they can be very different things.