Find and replace subarray in python array

Question

I have a 6600X5100 numpy array which represents a black & white image. I want to clear this image from black pixels noise- remove all black pixle lines (vertically and horizontally) that are shorter than 2 pixels.

So if I have something like this:

[0,  0,  0,   0,   0, 255]
[0, 255,255, 255, 255, 0 ]
[0, 255,255, 255,  0,  0 ]
[0, 255,255 ,255,  0, 255]
[0, 255,255, 255,  0, 255]
[0,  0,  0,   0,   0,  0 ]

The output array will be like this:

[0,  0,  0,   0,   0,  0 ]
[0, 255,255, 255,  0 , 0 ]
[0, 255,255, 255,  0,  0 ]
[0, 255,255 ,255,  0,  0 ]
[0, 255,255, 255,  0,  0 ]
[0,  0,  0,   0,   0,  0 ]

Performance is critical here so a simple loop over the array won't do. Is there a way to quickly find and replace subarray inside an array? So if [0, 255, 255, 0] or [0, 255, 0] is in the image array, replace those parts with 0.

Or if you have a better solution for this task, I will be grateful.

Is using scikit or scipy functions acceptable?

Dev-iL
– Dev-iL

2018-07-02 15:43:10 +00:00
Commented Jul 2, 2018 at 15:43 — Dev-iL
– Dev-iL, Commented Jul 2, 2018 at 15:43
scikit is fine

Phoenix
– Phoenix

2018-07-02 15:50:28 +00:00
Commented Jul 2, 2018 at 15:50 — Phoenix
– Phoenix, Commented Jul 2, 2018 at 15:50
Possible duplicate of Searching a sequence in a NumPy array

RedEyed
– RedEyed

2018-07-02 15:58:19 +00:00
Commented Jul 2, 2018 at 15:58 — RedEyed
– RedEyed, Commented Jul 2, 2018 at 15:58

James · Accepted Answer · 2018-07-02 16:21:51Z

4

You may want to look at the morphological filters of scikit-image.

You can define simple filters and use the opening function to clean up the image. You will have to play with the filters to get them exactly as you need them, but the library is very fast.

import numpy as np
from skimage.morphology import opening

img = np.array([[0,  0,  0,   0,   0, 255],
                [0, 255,255, 255, 255, 0 ],
                [0, 255,255, 255,  0,  0 ],
                [0, 255,255 ,255,  0, 255],
                [0, 255,255, 255,  0, 255],
                [0,  0,  0,   0,   0,  0 ]])


# horizontal and vertical filters
hf = np.array([[0,0,0,0,0],
               [0,1,1,1,0],
               [0,0,0,0,0]])
vf = hf.T

# apply each filter in turn
out = opening(opening(img, hf),vf)

out
# returns:
array([[  0,   0,   0,   0,   0,   0],
       [  0, 255, 255, 255,   0,   0],
       [  0, 255, 255, 255,   0,   0],
       [  0, 255, 255, 255,   0,   0],
       [  0, 255, 255, 255,   0,   0],
       [  0,   0,   0,   0,   0,   0]])

answered Jul 2, 2018 at 16:21

James

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Phoenix Over a year ago

Thanks. I used it with 'closing' instead of 'opening' since 255 represents white in my image.

Dev-iL · Accepted Answer · 2018-07-02 17:22:28Z

2

My solution is similar to the existing one, but I use 2d-convolutions:

import numpy as np
from scipy.signal import convolve2d as conv2

in_arr = np.array([
    [0,  0,  0,   0,   0, 255],
    [0, 255,255, 255, 255, 0 ],
    [0, 255,255, 255,  0,  0 ],
    [0, 255,255 ,255,  0, 255],
    [0, 255,255, 255,  0, 255],
    [0,  0,  0,   0,   0,  0 ]])

padded = np.pad(in_arr, 1, mode='constant', constant_values=0)

# Create a kernel
kern = np.ones((1, 3))

# Perform convolution
mask = np.logical_and((conv2(in_arr, kern,   mode='same') // 255) >= 2,
                      (conv2(in_arr, kern.T, mode='same') // 255) >= 2)

# Apply mask:
out_arr = in_arr * mask

Which also yields the desired result.

answered Jul 2, 2018 at 17:22

Dev-iL

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1 Comment

RedEyed Over a year ago

Really interesting solution!

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