The first step is to replace all the pixels in the image region with LBP (local binary pattern) codes. For every pixel, draw an imaginary circle of some radius centred on the pixel and draw some number of imaginary dots on that circle that are equally spaced apart. The radius and number of dots are to be hand tuned in order to maximise performance but they are usually set to a radius of 1 pixel and a number of dots of 8. Now pick the pixels that fall under those dots together with the pixel at the centre.

In the surrounding pixels, any pixel that has a lower intensity than the central pixel gets a zero and all the rest get a one. These numbers are put together in order to form a binary number called the LBP code of the central pixel.

Do this for every pixel in the region and then count how many times each code appears in the region, putting all the frequencies in a vector (a histogram). This is your texture descriptor and you can stop here. The more similar the vector is to other vectors, the more similar the texture of the image regions are.

We can do better than this however by making the codes uniform. If you do an analysis of these LBP code frequencies, you'll find that most of them have a most two bit changes in them. For example, 00011100 has two bit changes, one from a 0 to a 1 and another after from a 1 to a 0. On the other hand 01010101 has 7 changes. The reason why the majority will have at most two bit changes is because, most of the time, the pixels around a central pixel do not have random intensities but change in a flow as shown in the figure above. If there is a gradient of intensity changes in a single direction then there will only be two bit changes. This means that we can replace all LBP codes with more than two bit changes with a single LBP code that just means 'other'. This is called a uniform LBP code and it reduces our vector size significantly without much of a performance hit, which is good.

We can do even better by making the LBP codes rotation invariant, that is, using the same LBP code regardless of where the gradient direction is pointing. In the above figure, if the line of dark grey pixels were rotated, the LBP code would be changed to something like 00011110 for example or 11110000. The only thing that would change when rotating the surrounding pixels would be the bit rotation of the LBP code. We can make our codes invariant by artificially rotating our LBP codes so that for a given number of ones and zeros in the code we always have a canonical code. Usually we treat the binary code as an integer and rotate the bits until we get the minimum integer. For example, here are all the bit rotations of four 1s and four 0s:

11110000 (240), 01111000 (120), 00111100 (60), 00011110 (30), 00001111 (15)

In this case we would pick 00001111 as our canonical representation as that is the smallest integer. This drastically reduces our vector sizes as the amount of different possible LBP codes to find the frequency of will be a small fraction of the original full set of LBP codes.

You can use LBP descriptors using Python's skimage and numpy as follows:

import numpy as np import skimage import skimage.feature image = skimage.io.imread('img.png', as_gray=True) region = image[0:100,0:100] lbp_codes = skimage.feature.local_binary_pattern(region, P=8, R=1, method='uniform') #8 dots, radius 1, rotation invariant uniform codes. descriptor = np.histogram(lbp_codes, bins=10, range=(0, 10))[0] #Number of bins and range changes depending on the number of dots and LBP method used.

The number of bins and range needed can be obtained by artificially generating all the possible arrangements of darker and lighter pixels around a central pixel:

def get_histogram_params(dots, method): all_possible_lbp_codes = set() for i in range(2**8): str_bin = '{:0>8b}'.format(i) region = [ {'0':0,'1':2}[ch] for ch in str_bin ] #Surrounding pixels. region.insert(4, 1) #Central pixel. region = np.array(region).reshape((3,3)) lbp_codes = skimage.feature.local_binary_pattern(region, dots, 1, method) #Radius is minimum to keep region size to a minimum. all_possible_lbp_codes.add(lbp_codes[1,1]) #Get code of central pixel and add it to the set (keeps on unique values). num_bins = len(all_possible_lbp_codes) rng = (min(all_possible_lbp_codes), max(all_possible_lbp_codes)+1) return (num_bins, rng)

You can get more information about the skimage function from here.