CODEWITHSUNDEEP
pythonnumpyconvolution
Dominik Lemberger
Dominik Lemberger

Reputation: 2426

Convolution bluring image - python

I have this code below which is almost working - only thing I need here is the output of the convolution has to be devided by 9 and rounded down. Is this somehow possible with convolve2d?

import scipy
import scipy.signal
import numpy as np

def boxBlur(image):
    matrix = np.array(image)
    W = np.array([[1, 1, 1],
              [1, 1, 1],
              [1, 1, 1]])
    np.pad(matrix, 1, mode='constant')
    return scipy.signal.convolve2d(matrix, W, mode='valid')

So for this example:

boxBlur([[1,1,1],[1,7,1],[1,1,1]])

The output right now is [[15]] but it should be [[1]] (15/9=1,6666 rounded down=1)

Is there a way to not only use the convolve image on the matrix, but also do something else.

Right now the way I managed the problem is to manually go over the array and devide every cell by 9 rounding it down

Upvotes: 2

Views: 1253

Answers (1)

Divakar
Divakar

Reputation: 221564

That's called uniform filtetring, hence use SciPy's uniform_filter, which should be faster as well -

from scipy.ndimage import uniform_filter

uniform_filter(image.astype(float))[1:-1,1:-1]

Sample run -

In [38]: np.random.seed(0)
    ...: image = np.random.randint(0,9,(7,7))

In [39]: boxBlur(image)/9.0
Out[39]: 
array([[4.55555556, 5.        , 5.55555556, 5.44444444, 5.11111111],
       [4.44444444, 5.        , 5.        , 4.88888889, 4.22222222],
       [4.33333333, 4.44444444, 3.44444444, 3.44444444, 3.77777778],
       [2.22222222, 2.55555556, 2.88888889, 3.44444444, 3.55555556],
       [2.44444444, 2.11111111, 2.44444444, 3.55555556, 4.33333333]])

In [40]: uniform_filter(image.astype(float))[1:-1,1:-1]
Out[40]: 
array([[4.55555556, 5.        , 5.55555556, 5.44444444, 5.11111111],
       [4.44444444, 5.        , 5.        , 4.88888889, 4.22222222],
       [4.33333333, 4.44444444, 3.44444444, 3.44444444, 3.77777778],
       [2.22222222, 2.55555556, 2.88888889, 3.44444444, 3.55555556],
       [2.44444444, 2.11111111, 2.44444444, 3.55555556, 4.33333333]])

Timings -

In [42]: np.random.seed(0)
    ...: image = np.random.randint(0,9,(7000,7000))

In [43]: %timeit boxBlur(image)/9.0
1 loop, best of 3: 2.11 s per loop

In [44]: %timeit uniform_filter(image.astype(float))[1:-1,1:-1]
1 loop, best of 3: 612 ms per loop

Rounding down

For rounding it down, with the original solution, it would be : boxBlur(image)//9. The equivalent one here would be with floor-ing, so use np.floor(), but that might have precision issues. So, we might instead use np.round with a given number of decimal places for the precision and then floor with .astype(int) -

n = 10 # number of decimal places for precision
np.around(uniform_filter(image.astype(float))[1:-1,1:-1], decimals=n).astype(int)

For an input with ints, another way could be to scale up by 9 and round and then floor -

np.round(uniform_filter(image.astype(float))[1:-1,1:-1]*9)//9

Upvotes: 1

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