CODEWITHSUNDEEP
pythonnumpyrandompadding
Nick Nagy
Nick Nagy

Reputation: 61

Pad a numpy array with random values within a given range?

I want to be able to border an array (really, an image file) with a set of random values. The numpy.pad() function does not have any mode to do this. Is there a shorthand way to achieve this or would I have to create a function from scratch?

Upvotes: 3

Views: 1871

Answers (2)

kuppern87
kuppern87

Reputation: 1135

I think you need to create a padding function yourself to pass to np.pad. This one pads with random integers.

def random_pad(vec, pad_width, *_, **__):
    vec[:pad_width[0]] = np.random.randint(20, 30, size=pad_width[0])
    vec[vec.size-pad_width[1]:] = np.random.randint(30,40, size=pad_width[1])

You can use this with np.pad like this:

In [13]: img = np.arange(12).reshape(3, 4)

In [14]: img
Out[14]: 
array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11]])

In [15]: np.pad(img, ((2,3), (1,4)), mode=random_pad)
Out[15]: 
array([[26, 21, 22, 24, 21, 37, 37, 37, 39],
       [26, 25, 23, 29, 20, 39, 38, 30, 31],
       [26,  0,  1,  2,  3, 37, 31, 32, 36],
       [29,  4,  5,  6,  7, 30, 32, 33, 37],
       [24,  8,  9, 10, 11, 33, 34, 33, 37],
       [26, 36, 36, 36, 30, 32, 36, 38, 31],
       [29, 33, 34, 38, 35, 31, 33, 37, 33],
       [23, 37, 33, 33, 34, 32, 37, 33, 35]])

Upvotes: 2

Brad Solomon
Brad Solomon

Reputation: 40888

It may not be the most space-efficient, but one way to do this is to create a new array and place your existing array in the center of that.

>>> import numpy as np
>>> np.random.seed(444)
>>> arr = np.zeros((4, 5))  # your image array
>>> newsize = tuple(i + 2 for i in arr.shape)
>>> new = np.random.randint(low=0, high=50, size=newsize)
>>> new[1:-1, 1:-1] = arr
>>> new
array([[ 3, 48, 23,  8,  3, 39, 12],
       [47,  0,  0,  0,  0,  0, 15],
       [34,  0,  0,  0,  0,  0,  0],
       [ 6,  0,  0,  0,  0,  0,  6],
       [39,  0,  0,  0,  0,  0, 13],
       [ 2, 15,  9, 34,  9, 24, 25]])

This is assuming your image is grayscale (2d) rather than a 3-dimensional MxNx4 RGBA array. In that case, you'd want new[1:-1, 1:-1, 1:-1].

You can also do this by passing a callable to np.pad(), but there is a caveat (see below):

from functools import partial

def _pad_random(vector, pad_width, iaxis, kwargs, low, high):
    a, b = np.random.randint(low, high, size=2)
    vector[:pad_width[0]] = a
    vector[-pad_width[1]:] = b
    return vector

pad_random = partial(_pad_random, low=0, high=50)

Usage:

>>> np.pad(arr, 1, pad_random)
array([[23., 19.,  6., 47., 17.,  7., 26.],
       [26.,  0.,  0.,  0.,  0.,  0., 37.],
       [39.,  0.,  0.,  0.,  0.,  0., 39.],
       [39.,  0.,  0.,  0.,  0.,  0., 42.],
       [28.,  0.,  0.,  0.,  0.,  0., 47.],
       [11., 32., 37.,  2., 38., 30., 44.]])

Caveat: it looks like, when you pass a function to the mode parameter of np.pad(), this function is called multiple times. Here's an example straight from the docs with a few print calls sprinkled in:

>>> def pad_with(vector, pad_width, iaxis, kwargs):
...     pad_value = kwargs.get('padder', 10)
...     print(vector[:pad_width[0]])
...     print(vector[-pad_width[1]:])
...     vector[:pad_width[0]] = pad_value
...     vector[-pad_width[1]:] = pad_value
...     return vector
...
>>> arr = np.arange(6).reshape(3, 2)

>>> np.pad(arr, 1, pad_with)
[0]
[0]
[0]
[0]
[0]
[0]
[0]
[0]
[10]
[10]
[0]
[0]
[0]
[0]
[0]
[0]
[10]
[10]
array([[10, 10, 10, 10],
       [10,  0,  1, 10],
       [10,  2,  3, 10],
       [10,  4,  5, 10],
       [10, 10, 10, 10]])

So, as long as the images aren't too big, it may be much more time efficient to use the first approach above.

Upvotes: 1

Related Questions