How to understand numpy strides for layman?

Question

I am currently going through numpy and there is a topic in numpy called \"strides\". I understand what it is. But how does it work? I did not find any useful information onl

Answer 1

Just to add to great answer by @AndyK, I learnt about numpy strides from Numpy MedKit. There they show the use with a problem as follows:

Given input:

x = np.arange(20).reshape([4, 5])
>>> x
array([[ 0,  1,  2,  3,  4],
       [ 5,  6,  7,  8,  9],
       [10, 11, 12, 13, 14],
       [15, 16, 17, 18, 19]])

Expected Output:

array([[[  0,  1,  2,  3,  4],
        [  5,  6,  7,  8,  9]],

       [[  5,  6,  7,  8,  9],
        [ 10, 11, 12, 13, 14]],

       [[ 10, 11, 12, 13, 14],
        [ 15, 16, 17, 18, 19]]])

To do this, we need to know the following terms:

shape - The dimensions of the array along each axis.

strides - The number of bytes of memory that must be skipped to progress to the next item along a certain dimension.

>>> x.strides
(20, 4)

>>> np.int32().itemsize
4

Now, if we look at the Expected Output:

array([[[  0,  1,  2,  3,  4],
        [  5,  6,  7,  8,  9]],

       [[  5,  6,  7,  8,  9],
        [ 10, 11, 12, 13, 14]],

       [[ 10, 11, 12, 13, 14],
        [ 15, 16, 17, 18, 19]]])

We need to manipulate the array shape and strides. The output shape must be (3, 2, 5), i.e. 3 items, each containing two rows (m == 2) and each row having 5 elements.

The strides need to change from (20, 4) to (20, 20, 4). Each item in the new output array starts at a new row, that each row consists of 20 bytes (5 elements of 4 bytes each), and each element occupies 4 bytes (int32).

So:

>>> from numpy.lib import stride_tricks
>>> stride_tricks.as_strided(x, shape=(3, 2, 5),
                                strides=(20, 20, 4))
...
array([[[  0,  1,  2,  3,  4],
        [  5,  6,  7,  8,  9]],

       [[  5,  6,  7,  8,  9],
        [ 10, 11, 12, 13, 14]],

       [[ 10, 11, 12, 13, 14],
        [ 15, 16, 17, 18, 19]]])

An alternative would be:

>>> d = dict(x.__array_interface__)
>>> d['shape'] = (3, 2, 5)
>>> s['strides'] = (20, 20, 4)

>>> class Arr:
...     __array_interface__ = d
...     base = x

>>> np.array(Arr())
array([[[  0,  1,  2,  3,  4],
        [  5,  6,  7,  8,  9]],

       [[  5,  6,  7,  8,  9],
        [ 10, 11, 12, 13, 14]],

       [[ 10, 11, 12, 13, 14],
        [ 15, 16, 17, 18, 19]]])

I use this method very often instead of numpy.hstack or numpy.vstack and trust me, computationally it is much faster.

Note:

When using very large arrays with this trick, calculating the exact strides is not so trivial. I usually make a numpy.zeroes array of the desired shape and get the strides using array.strides and use this in the function stride_tricks.as_strided.

Hope it helps!