[image] Checking images for similarity with OpenCV

Does OpenCV support the comparison of two images, returning some value (maybe a percentage) that indicates how similar these images are? E.g. 100% would be returned if the same image was passed twice, 0% would be returned if the images were totally different.

I already read a lot of similar topics here on StackOverflow. I also did quite some Googling. Sadly I couldn't come up with a satisfying answer.

If for matching identical images ( same size/orientation )

// Compare two images by getting the L2 error (square-root of sum of squared error).
double getSimilarity( const Mat A, const Mat B ) {
if ( A.rows > 0 && A.rows == B.rows && A.cols > 0 && A.cols == B.cols ) {
    // Calculate the L2 relative error between images.
    double errorL2 = norm( A, B, CV_L2 );
    // Convert to a reasonable scale, since L2 error is summed across all pixels of the image.
    double similarity = errorL2 / (double)( A.rows * A.cols );
    return similarity;
}
else {
    //Images have a different size
    return 100000000.0;  // Return a bad value
}

Source

Sam's solution should be sufficient. I've used combination of both histogram difference and template matching because not one method was working for me 100% of the times. I've given less importance to histogram method though. Here's how I've implemented in simple python script.

import cv2

class CompareImage(object):

    def __init__(self, image_1_path, image_2_path):
        self.minimum_commutative_image_diff = 1
        self.image_1_path = image_1_path
        self.image_2_path = image_2_path

    def compare_image(self):
        image_1 = cv2.imread(self.image_1_path, 0)
        image_2 = cv2.imread(self.image_2_path, 0)
        commutative_image_diff = self.get_image_difference(image_1, image_2)

        if commutative_image_diff < self.minimum_commutative_image_diff:
            print "Matched"
            return commutative_image_diff
        return 10000 //random failure value

    @staticmethod
    def get_image_difference(image_1, image_2):
        first_image_hist = cv2.calcHist([image_1], [0], None, [256], [0, 256])
        second_image_hist = cv2.calcHist([image_2], [0], None, [256], [0, 256])

        img_hist_diff = cv2.compareHist(first_image_hist, second_image_hist, cv2.HISTCMP_BHATTACHARYYA)
        img_template_probability_match = cv2.matchTemplate(first_image_hist, second_image_hist, cv2.TM_CCOEFF_NORMED)[0][0]
        img_template_diff = 1 - img_template_probability_match

        # taking only 10% of histogram diff, since it's less accurate than template method
        commutative_image_diff = (img_hist_diff / 10) + img_template_diff
        return commutative_image_diff


    if __name__ == '__main__':
        compare_image = CompareImage('image1/path', 'image2/path')
        image_difference = compare_image.compare_image()
        print image_difference

A little bit off topic but useful is the pythonic numpy approach. Its robust and fast but just does compare pixels and not the objects or data the picture contains (and it requires images of same size and shape):

A very simple and fast approach to do this without openCV and any library for computer vision is to norm the picture arrays by

import numpy as np
picture1 = np.random.rand(100,100)
picture2 = np.random.rand(100,100)
picture1_norm = picture1/np.sqrt(np.sum(picture1**2))
picture2_norm = picture2/np.sqrt(np.sum(picture2**2))

After defining both normed pictures (or matrices) you can just sum over the multiplication of the pictures you like to compare:

1) If you compare similar pictures the sum will return 1:

In[1]: np.sum(picture1_norm**2)
Out[1]: 1.0

2) If they aren't similar, you'll get a value between 0 and 1 (a percentage if you multiply by 100):

In[2]: np.sum(picture2_norm*picture1_norm)
Out[2]: 0.75389941124629822

Please notice that if you have colored pictures you have to do this in all 3 dimensions or just compare a greyscaled version. I often have to compare huge amounts of pictures with arbitrary content and that's a really fast way to do so.