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

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 (?)