integralImage
Calculate 2-D integral image
Description
In an integral image, each pixel represents the cumulative sum of a corresponding input pixel with all pixels above and to the left of the input pixel.
An integral image enables you to rapidly calculate summations over image subregions. Subregion summations can be computed in constant time as a linear combination of only four pixels in the integral image, regardless of the size of the subregion. Use of integral images was popularized by the Viola-Jones algorithm [1].
Examples
Create a simple sample matrix.
I = magic(5)
I = 5×5
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
Calculate the integral image of the sample matrix. These steps show how the first few values in the original matrix map to values in the integral image. Note that the pixel with (row, column) coordinate (r, c) in the original image corresponds to the pixel with coordinate (r+1, c+1) in the integral image.
The first row and column in the integral image are all
0s.The pixel in the original matrix at coordinate (1, 1) with value 17 is unchanged in the integral image because there are no other pixels in the summation. Therefore, the pixel in the integral image at coordinate (2, 2) has the value 17.
The pixel in the original matrix at coordinate (1, 2) maps to the pixel (2, 3) in the integral image. The value is the summation of the original pixel value (24), the pixels above it (0), and the pixels to its left (17): 24 + 17 + 0 = 41.
The pixel in the original matrix at coordinate (1, 3) maps to the pixel (2, 4) in the integral image. The value is the summation of the original pixel value (1), the pixel above it (0), and the pixels to its left (which have already been summed to 41). Thus the value at pixel (2,4) in the integral image is 1 + 41 + 0 = 42.
J = integralImage(I)
J = 6×6
0 0 0 0 0 0
0 17 41 42 50 65
0 40 69 77 99 130
0 44 79 100 142 195
0 54 101 141 204 260
0 65 130 195 260 325
Read a grayscale image into the workspace. Display the image.
I = imread('pout.tif');
imshow(I)
Compute the integral image.
J = integralImage(I);
Use the drawrectangle tool to select a rectangular subregion. The tool returns a Rectangle object.
d = drawrectangle;
The Vertices property of the Rectangle object stores the coordinates of vertices as a 4-by-2 matrix. Vertices are ordered starting with the top-left and continuing in a clockwise direction. Split the matrix into two vectors containing the row and column coordinates. Because the integral image is zero-padded on the top and left side, increment the row and column coordinates by 1 to retrieve the corresponding elements of the integral array.
r = floor(d.Vertices(:,2)) + 1; c = floor(d.Vertices(:,1)) + 1;
Calculate the sum of all pixels in the rectangular subregion by combining four pixels of the integral image.
regionSum = J(r(1),c(1)) - J(r(2),c(2)) + J(r(3),c(3)) - J(r(4),c(4))
regionSum = 613092
Create a simple sample matrix.
I = magic(5)
I = 5×5
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
Create an integral image with a rotated orientation.
J = integralImage(I,'rotated')J = 6×7
0 0 0 0 0 0 0
0 17 24 1 8 15 0
17 64 47 40 38 39 15
64 74 91 104 105 76 39
74 105 149 188 183 130 76
105 170 232 272 236 195 130
Define a rotated rectangular subregion. This example specifies a subregion with top corner at coordinate (1,3) in the original image. The subregion has a rotated height of 1 and width of 2.
r = 1; c = 3; h = 1; w = 2;
Get the value of the four corner pixels of the subregion in the integral image.
regionBottom = J(r+w+h,c-h+w+1); regionTop = J(r,c+1); regionLeft = J(r+h,c-h+1); regionRight = J(r+w,c+w+1); regionCorners = [regionBottom regionTop regionLeft regionRight]
regionCorners = 1×4
105 0 24 39
Calculate the sum of pixels in the subregion by summing the four corner pixel values.
regionSum = regionBottom + regionTop - regionLeft - regionRight
regionSum = 42
Input Arguments
Output Arguments
Algorithms
Every pixel in an integral image represents the summation of the
corresponding input pixel value with all input pixels above and to
the left of the input pixel. Because
integralImage zero-pads the resulting
integral image, a pixel with (row, column) coordinate
(m, n) in the original
image maps to the pixel with coordinate (m+1,
n+1) in the integral image.
In the figure, the current pixel in the input image is the dark green pixel at coordinate (4, 5). All pixels in the input image above and to the left of the input pixel are colored in light green. The summation of the green pixel values is returned in the integral image pixel with coordinate (5, 6), colored in gray.
integralImage performs a faster computation of the
integral image by summing pixel values in both the input image and
the integral image. Pixel (m,
n) in integral image
J is a linear combination of only four
pixels: one from the input image and three previously-calculated
pixels from the integral image.
J(m,n) = J(m,n-1) + J(m-1,n) +
I(m-1,n-1) - J(m-1,n-1)
This figure shows which pixels are included in the sum when calculating the integral image at the gray pixel. Green pixels add to the sum and red pixels subtract from the sum.
If you specify the image orientation as
"rotated", then pixels in an integral
image represent the summation of a corresponding input pixel value
with all input pixels that are diagonally above the input pixel.
integralImage performs the summation
along diagonal lines. This approach is less computationally
intensive than rotating the image and calculating the integral image
in rectilinear directions.
In the figure, the current pixel in the input image is the dark green pixel at coordinate (4, 5). All pixels in the input image diagonally above the input pixel are colored in light green. The summation of the green pixel values is returned in the integral image pixel with coordinate (5, 6), colored in gray.
integralImage performs a faster computation of the
rotated integral image by summing pixel values in both the input
image and the integral image. Pixel (m,
n) in integral image
J is a linear combination of only five
pixels: two from the input image and three previously-calculated
pixels from the integral image:
J(m,n) = J(m-1,n-1) + J(m-1,n+1) - J(m-2,n) +
I(m-1,n-1) + I(m-2,n-1)
This figure shows which pixels are included in the sum when calculating the integral image at the gray pixel. Green pixels add to the sum and red pixels subtract from the sum.
A subregion in an upright orientation with top-left coordinate
(m,n), height
h, and width w in the
original image has the summation:
regionSum = J(m–1,n–1) + J(m+h–1,n+w–1)
– J(m+h–1,n–1) –
J(m-1,n+w-1)
For example, in the input image below, the summation of the blue shaded region is: 46 – 22 – 20 + 10 = 14. The calculation subtracts the regions above and to the left of the shaded region. The area of overlap is added back to compensate for the double subtraction.
A subregion in an rotated orientation uses a different definition of height and width [2]. The summation of the region is:
regionSum = J(m+h+w,n-h+w+1) + J(m,n+1) -
J(m+h,n-h+1) - J(m+w,n+w+1)
References
[1] Viola, P., and M. J. Jones. "Rapid Object Detection using a Boosted Cascade of Simple Features". Proceedings of the 2001 IEEE Computer Society Conference on Computer Vision and Pattern Recognition. 2001. Vol. 1, pp. 511–518.
[2] Lienhart, R., and J. Maydt. "An Extended Set of Haar-like Features for Rapid Object Detection". Proceedings of the 2002 IEEE International Conference on Image Processing. Sept. 2002. Vol. 1, pp. 900–903.
