CODEWITHSUNDEEP
pythonarraysnumpytransformationedge-detection
MarkD
MarkD

Reputation: 4944

2D numpy array- check to see if all adjacent terms are equal

I am starting with a nxm boolean array, that defines regions- true if it is in the region, false if not. For example:

r = np.array([[ 0,  0,  1,  1,  1],
              [ 0,  1,  1,  0,  0],
              [ 0,  1,  1,  1,  0],
              [ 1,  1,  1,  0,  0],
              [ 1,  1,  0,  0,  0]])

The line between regions can be defined as an n-1 x m-1 array, which would represent a 'point' in between each set of four values. If all 4 surrounding values are the same, you are not on the edge between regions. If any of the 4 values are different, you are. For r above:

l = np.array([[ 1, 1, 1, 1],
              [ 1, 0, 1, 1],
              [ 1, 0, 1, 1],
              [ 0, 1, 1, 0]])

Any thoughts on how this can be done efficiently? I have tried doing a diff in both directions, but this doubles up. Is there a 2D diff function? Some other way of doing this?

Upvotes: 2

Views: 2134

Answers (2)

tom10
tom10

Reputation: 69182

An good answer's already selected, but I like solutions that are easy to write and understand so I'll still post this:

from scipy.signal import convolve2d
kernel = np.array(((1, 1), (1, 1)))

x = convolve2d(r, kernel, mode="valid")   # sum all the neighboring values (and mode handles the boundary issues)
x[x==4] = 0                               # set the elements that sum to 4 to zero (the ones that sum to 0 are already 0)
x[x>0] = 1                                # anything greater than one is set to 1

[[1 1 1 1]
 [1 0 1 1]
 [1 0 1 1]
 [0 1 1 0]]

Upvotes: 2

farenorth
farenorth

Reputation: 10781

This will do the test for points surrounded by True,

tmp = r[1:] & r[:-1]
l = np.logical_not(tmp[:, 1:] & tmp[:, :-1])

Then you can do the test for points surrounded by False the same way and combine them,

r = np.logical_not(r)
tmp = r[1:] & r[:-1]
l &= np.logical_not(tmp[:, 1:] & tmp[:, :-1])

print l.astype(int)
# [[1 1 1 1]
#  [1 0 1 1]
#  [1 0 1 1]
#  [0 1 1 0]]

Upvotes: 6

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