Computing image integral

Question

How do one find the mean, std dev, and gradient from image integral? Given an image such as follows:

Answer 1

What jmch didn't say is, if sqrt( C'+A'-B'-D'/K - (mean*mean) ) is not how you compute the standard deviation from the integral image, then how you do it?

First, let me switch to Python / numpy code, so we get a modicum of notation consistency and expressions are easier to check. Given a sample array X, say:

X = array([random() * 10.0 for i in range(0, 9)])

The uncorrected sample standard deviation of X can be defined as:

std = (sum((X - mean(X)) ** 2) / len(X)) ** 0.5 # 1

Applying the binomial theorem to (X - mean(X)) ** 2 we get:

std = (sum(X ** 2 - X * 2 * mean(X) + mean(X) ** 2) / len(X)) ** 0.5 # 2

Given the identities of the summation operation, we can make:

std = ((sum(X ** 2) - 2 * mean(X) * sum(X) + len(X) * mean(X) ** 2) / len(X)) ** 0.5 # 3

If we make S = sum(X), S2 = sum(X ** 2), M = mean(X) and N = len(X) we get:

std = ((S2 - 2 * M * S + N * M ** 2) / N) ** 0.5 # 4

Now for an image I and two integral images P and P2 calculated from I (where P2 is the integral image for squared pixel values), we know that, given the four edge coordinates A = (i0, j0), B = (i0, j1), C = (i1, j0) and D = (i1, j1), the values of S, S2, M and N can be calculated for the range I[A:D] as:

S = P[A] + P[D] - P[B] - P[C]

S2 = P2[A] + P2[D] - P2[B] - P2[C]

N = (i1 - i0) * (j1 - j0)

M = S / N

Which can then be applied to equation (4) above yielding the standard deviation of the range I[A:D].

Edit: It's not entirely necessary, but given that M = S / N we can apply the following substitutions and simplifications to equation (4):

std = ((S2 - 2 * M * S + N * M ** 2) / N) ** 0.5

std = ((S2 - 2 * (S / N) * S + N * (S / N) ** 2) / N) ** 0.5

std = ((S2 - 2 * ((S ** 2) / N) + (S ** 2 / N)) / N) ** 0.5

std = ((S2 - ((S ** 2) / N)) / N) ** 0.5

std = (S2 / N - (S / N) ** 2) ** 0.5 # 5

Which is quite close to the equation remi gave, actually.

Answer 2

C+A-B-D gives you the sum of the gray levels in the zone delimited by A,B,C,D, so, to get the mean you just need to dived it by the area of the zone:

mean = (C+A-B-D)/4

To get the dev, you must compute the sum of square area table (using cv::integral you can pass a additional parameters to get the sum of squares). Quoting wikipedia, the standard deviation is equal to the square root of (the average of the squares less the square of the average). So assuming A',B',C',D' the values in your square area table:

dev = sqrt((C'+A'-B'-D')/4 - (mean*mean))

So computing mean and dev using integral image is very fast using integral images, especially if you want to compute those quantities at random locations and on random size of image patches.

Concerning the gradient, it's more complex. Are you sure you do not want to use sobel operator?

Answer 3

If C+A-B-C is the sum of all gray levels in the zone, then the mean is not

mean = C+A-B-D/4

but

mean = C+A-B-D/K

where K is the number of graylevels in the zone.

Also,

dev = sqrt( C'+A'-B'-D'/4 - (mean*mean) )

is not the stdev, because

dev = sqrt( (1/N)*sum_N ( x_i - u )^2 )

the equation here is equivalent to

dev = sqrt( (1/N)*sum_N ( (x_i)^2 ) - u^2 )

those equations are not equivalent.