The example come with alternative solution: Histogram method.
The script was run under Fedora 25.
If the images are the same the result will be 0.0.
For testing I change the image2.png by make a line into this with a coverage of 10%.
The result of the script was:
1116.63243729
The images come with this dimensions: 738 x 502 px.
import math
import operator
from math import *
import PIL
from PIL import Image
h1 = Image.open("image1.png").histogram()
h2 = Image.open("image2.png").histogram()
rms = math.sqrt(reduce(operator.add,
map(lambda a,b: (a-b)**2, h1, h2))/len(h1))
print rms
Example:
operator.lt(a, b)
operator.le(a, b)
operator.eq(a, b)
operator.ne(a, b)
operator.ge(a, b)
operator.gt(a, b)
operator.__lt__(a, b)
operator.__le__(a, b)
operator.__eq__(a, b)
operator.__ne__(a, b)
operator.__ge__(a, b)
operator.__gt__(a, b)This is like math operators:
lt(a, b) is equivalent to a < b
le(a, b) is equivalent to a <= b
Another example:
>>> # Elementwise multiplication
>>> map(mul, [0, 1, 2, 3], [10, 20, 30, 40])
[0, 20, 60, 120]
>>> # Dot product
>>> sum(map(mul, [0, 1, 2, 3], [10, 20, 30, 40]))
200
