Implementation of the Bhattacharyya distance in Python

bhattacharyya

# bhattacharyya test
import numpy
import math

h1 = [ 1, 2, 3, 4, 5, 6, 7, 8 ];
h2 = [ 6, 5, 4, 3, 2, 1, 0, 0 ];
h3 = [ 8, 7, 6, 5, 4, 3, 2, 1 ];
h4 = [ 1, 2, 3, 4, 4, 3, 2, 1 ];
h5 = [ 8, 8, 8, 8, 8, 8, 8, 8 ];

h = [ h1, h2, h3, h4, h5 ];

def mean( hist ):
    mean = 0.0;
    for i in hist:
        mean += i;
    mean/= len(hist);
    return mean;

def bhatta ( hist1,  hist2):
    # calculate mean of hist1
    h1_ = mean(hist1);

    # calculate mean of hist2
    h2_ = mean(hist2);

    # calculate score
    score = 0;
    for i in range(8):
        score += math.sqrt( hist1[i] * hist2[i] );
    # print h1_,h2_,score;
    score = math.sqrt( 1 - ( 1 / math.sqrt(h1_*h2_*8*8) ) * score );
    return score;

# generate and output scores
scores = [];
for i in range(len(h)):
    score = [];
    for j in range(len(h)):
        score.append( bhatta(h[i],h[j]) );
    scores.append(score);

for i in scores:
    print i

Seeing as you import numpy, you might as well use its mean function. np.average(hist).

Why not directly convert the hist1, hist2 to the percentage by dividing the sum of each instead of calculating the mean, then divide by the mean * 8?

Very useful. I have a quiestion. Why you do the for in range of 8? 8 is the size of each histogram? if this is the case, can i change 8 by len(h1) for example?. Thanks

def bhattacharyya(h1, h2):
  '''Calculates the Byattacharyya distance of two histograms.'''

  def normalize(h):
    return h / np.sum(h)

  return 1 - np.sum(np.sqrt(np.multiply(normalize(h1), normalize(h2))))

if we want to use bhattacharyya distance for an image with more number of bands ( which will be a 3d numpy array) what modifications we have to do in order to use above code for that image.

You implemented Hellinger distance which is different from Bhattacharyya distance.

@harry098 maybe using flatten so your array will be 1D array (?)