Compare two images the python/linux way

There is a OSS project that uses WebDriver to take screen shots and then compares the images to see if there are any issues (http://code.google.com/p/fighting-layout-bugs/)). It does it by openning the file into a stream and then comparing every bit.

You may be able to do something similar with PIL.

EDIT:

After more research I found

h1 = Image.open("image1").histogram()
h2 = Image.open("image2").histogram()

rms = math.sqrt(reduce(operator.add,
    map(lambda a,b: (a-b)**2, h1, h2))/len(h1))

on http://snipplr.com/view/757/compare-two-pil-images-in-python/ and http://effbot.org/zone/pil-comparing-images.htm

I guess you should decode the images and do a pixel by pixel comparison to see if they're reasonably similar.

With PIL and Numpy you can do it quite easily:

import Image
import numpy
import sys

def main():
    img1 = Image.open(sys.argv[1])
    img2 = Image.open(sys.argv[2])

    if img1.size != img2.size or img1.getbands() != img2.getbands():
        return -1

    s = 0
    for band_index, band in enumerate(img1.getbands()):
        m1 = numpy.array([p[band_index] for p in img1.getdata()]).reshape(*img1.size)
        m2 = numpy.array([p[band_index] for p in img2.getdata()]).reshape(*img2.size)
        s += numpy.sum(numpy.abs(m1-m2))
    print s

if __name__ == "__main__":
    sys.exit(main())

This will give you a numeric value that should be very close to 0 if the images are quite the same.

Note that images that are shifted/rotated will be reported as very different, as the pixels won't match one by one.

Using ImageMagick, you can simply use in your shell [or call via the OS library from within a program]

compare image1 image2 output

This will create an output image with the differences marked

compare -metric AE -fuzz 5% image1 image2 output

Will give you a fuzziness factor of 5% to ignore minor pixel differences. More information can be procured from here

the problem of knowing what makes some features of the image more important than other is a whole scientific program. I would suggest some alternatives depending on the solution you want:

• if your problem is to see if there is a flipping of bits in your JPEGs, then try to image the difference image (there was perhaps a minor edit locally?),

• to see if images are globally the same, use the Kullback Leibler distance to compare your histograms,

• to see if you have some qualittative change, before applying other answers, filter your image using the functions below to raise the importance of high-level frequencies:

code:

def FTfilter(image,FTfilter):
    from scipy.fftpack import fft2, fftshift, ifft2, ifftshift
    from scipy import real
    FTimage = fftshift(fft2(image)) * FTfilter
    return real(ifft2(ifftshift(FTimage)))
    #return real(ifft2(fft2(image)* FTfilter))


#### whitening
def olshausen_whitening_filt(size, f_0 = .78, alpha = 4., N = 0.01):
    """
    Returns the whitening filter used by (Olshausen, 98)

    f_0 = 200 / 512

    /!\ you will have some problems at dewhitening without a low-pass

    """
    from scipy import mgrid, absolute
    fx, fy = mgrid[-1:1:1j*size[0],-1:1:1j*size[1]]
    rho = numpy.sqrt(fx**2+fy**2)
    K_ols = (N**2 + rho**2)**.5 * low_pass(size, f_0 = f_0, alpha = alpha)
    K_ols /= numpy.max(K_ols)

    return  K_ols

def low_pass(size, f_0, alpha):
    """
    Returns the low_pass filter used by (Olshausen, 98)

    parameters from Atick (p.240)
    f_0 = 22 c/deg in primates: the full image is approx 45 deg
    alpha makes the aspect change (1=diamond on the vert and hor, 2 = anisotropic)

    """

    from scipy import mgrid, absolute
    fx, fy = mgrid[-1:1:1j*size[0],-1:1:1j*size[1]]
    rho = numpy.sqrt(fx**2+fy**2)
    low_pass = numpy.exp(-(rho/f_0)**alpha)

    return  low_pass

(shameless copy from http://www.incm.cnrs-mrs.fr/LaurentPerrinet/Publications/Perrinet08spie )

Using only PIL and some of the Python math libraries, it is possible to see if two images are identical to each other in a simple, concise manner. This method has only been tested on image files with the same dimensions and extension, but has avoided several errors made in other answers to this question.

import math, operator
from PIL import Image
from PIL import ImageChops

def images_are_similar(img1, img2, error=90):
    diff = ImageChops.difference(img1, img2).histogram()
    sq = (value * (i % 256) ** 2 for i, value in enumerate(diff))
    sum_squares = sum(sq)
    rms = math.sqrt(sum_squares / float(img1.size[0] * img1.size[1]))

    # Error is an arbitrary value, based on values when 
    # comparing 2 rotated images & 2 different images.
    return rms < error

Advantages:
Adding % 256 to the computation of squares weights each color equally. Many previous answers' RMS formulas give Blue pixel values 3x the weight of Red values, and Green pixel values 2x the weight of Red values.

Easier to grok. While the RMS calculation could be written as a one-liner, with lambdas and the reduce method, expanding it out to 3 lines greatly improves at-a-glance readability.

This code properly detects that rotated images are different from a differently oriented base image. This avoids a pitfall when using histograms to compare images, as pointed out by @musicinmybrain. If histograms of 2 images are created then compared to each other, if one image is a rotation of the other, the comparison will report that there are no differences in the images because the images' histograms are identical. On the other hand, if the images are compared first, then a histogram of the comparison results is created, the images will compare accurately, even if one is a rotation of the other.

The code used in this answer is a copy/paste from this code.activestate.com post, taking into consideration the 3rd comment, which corrects the heavier weighting of Green and Blue pixel values.

First, I should note they’re not identical; b has been recompressed and lost quality. You can see this if you look carefully on a good monitor.

To determine that they are subjectively “the same,” you would have to do something like what fortran suggested, although you will have to arbitrarily establish a threshold for “sameness.” To make s independent of image size, and to handle channels a little more sensibly, I would consider doing the RMS (root mean square) Euclidean distance in colorspace between the pixels of the two images. I don’t have time to write out the code right now, but basically for each pixel, you compute

(R_2 - R_1) ** 2 + (G_2 - G_1) ** 2 + (B_2 - B_1) ** 2

, adding in an

(A_2 - A_1) ** 2

term if the image has an alpha channel, etc. The result is the square of the colorspace distance between the two images. Find the mean (average) across all pixels, then take the square root of the resulting scalar. Then decide a reasonable threshold for this value.

Or, you might just decide that copies of the same original image with different lossy compression are not truly “the same” and stick with the file hash.

You can either compare it using PIL (iterate through pixels / segments of the picture and compare) or if you're looking for a complete identical copy comparison, try comparing the MD5 hash of both files.

A simple solution with numpy

img1 = Image.open(img1_path)
img2 = Image.open(img2_path)

Then use array_equal

np.array_equal(img1, img2)

If the return is True then their exactly the same across all channels

import cv2
import numpy as np

original = cv2.imread("Test.jpg")
duplicate = cv2.imread("Test1.jpg")

# 1) Check if 2 images are equals
if original.shape == duplicate.shape:
    print("The images have same size and channels")
    difference = cv2.subtract(original, duplicate)
    b, g, r = cv2.split(difference)


    if cv2.countNonZero(b) == 0 and cv2.countNonZero(g) == 0 and cv2.countNonZero(r) == 0:
        print("The images are completely Equal")
        
cv2.imshow("Original", original)
cv2.imshow("Duplicate", duplicate)
cv2.waitKey(0)
cv2.destroyAllWindows()

if you want to check two images same or not,this code may help you.

 import binascii
 
 
 with open('pic1.png', 'rb') as f:
     content1 = f.read()
 with open('pic2.png', 'rb') as f:
     content2 = f.read()
 if content1 == content2:
     print("same")
 else:
     print("not same")

`