Create buffer zone within a Numpy array

Question

I have a binary image as follows:

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

For pixels having 1s values, I want to make buffer zone of two pixels with value 1s surrounded in every four directions. The expected result would be:

result=np.array([[1, 1 , 1 , 1 , 1 , 1 , 1 , 1],
                 [1, 1 , 1 , 1 , 1 , 1 , 1 , 1],
                 [1, 1 , 1 , 1 , 1 , 1 , 1 , 1],
                 [1, 1 , 1 , 1 , 1 , 1 , 1 , 1],
                 [1, 1 , 1 , 1 , 1 , 1 , 1 , 1],
                 [1, 1 , 1 , 1 , 1 , 1 , 1 , 1],
                 [1, 1 , 1 , 1 , 1 , 1 , 1 , 1]])

How can I do it?

Is it an array of only ones and zeros, or are there other values? — Henry Gomersall
– Henry Gomersall, Commented Mar 13, 2015 at 8:39

Henry Gomersall · Accepted Answer · 2015-03-13 08:46:04Z

2

If you only have ones and zeros on the input and output array, you can do it with a 2D convolution, which is simple and works.

from scipy.signal import convolve2d

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

# the kernel doesn't need to be ones, it just needs to be positive and
# non-zero.
kernel = np.ones((5, 5))

result = np.int64(convolve2d(data, kernel, mode='same') > 0)

Which gives you the output you want. You need to define what you want to happen at the edges - in this version, the output array is the same size as the input array.

It might be possible you can do something faster if you have a sparse array.

If you have other values than one and zero in your array, more thought would be needed.

answered Mar 13, 2015 at 8:46

Henry Gomersall

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8 Comments

Eric O. Lebigot Over a year ago

Why would you need "a positive and non-zero kernel"? I can't see any reason why, and the documentation does not mention this either (docs.scipy.org/doc/scipy/reference/generated/…).

Henry Gomersall Over a year ago

You don't for the convolve2d, but you do for the method to work (specifically, the > 0).

Borys Over a year ago

@HenryGomersall Thanks for your answer. Could you explain me why you have used kernel of (5,5).

Henry Gomersall Over a year ago

That creates a 5x5 array of ones. Are you familiar with convolution? It's probably a bit involved to explain in a comment, but in essence, the 5x5 shape is what you desire at the output for a single 1 in the input.

Henry Gomersall Over a year ago

It's worth mentioning that the 2D kernel is separable, so you can do this with a pair of 1D kernels along each dimension, which would be faster if that's an issue.

|

Imanol Luengo · Accepted Answer · 2015-04-08 15:27:53Z

You can also do it using morphological dilation operator (which dilates the ones in this case).

from skimage.morphology import square, dilation
data = np.array([[0, 0 , 0 , 0 , 0 , 0 , 0 , 0],
                 [0, 0 , 0 , 0 , 0 , 0 , 0 , 0],
                 [0, 0 , 1 , 1 , 1 , 1 , 0 , 0],
                 [0, 0 , 1 , 1 , 1 , 1 , 0 , 0],
                 [0, 0 , 1 , 1 , 1 , 1 , 0 , 0],
                 [0, 0 , 0 , 0 , 0 , 0 , 0 , 0],
                 [0, 0 , 0 , 0 , 0 , 0 , 0 , 0]])
result = dilation(data, square(5))

Note that square(5) is equivalent to np.ones((5,5)) in this case. Dilation operator works by dilating True or 1 pixels with the element passed as a second parameter (in this case, a 5x5 square centered at each pixel).

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