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# Feature Detection Test for Fish Classification

Module by: Kyle Clarkson, Jason Sedano, Ian Clark. E-mail the authors

Summary: This test determines whether a fish is a salmon or a trout based on different color features that are detected using the 2-D DWT

One of the most important features of different fish are the colors of their heads, tails, and bodies. This test breaks down the different color matricies into blocks of similar colors and uses them to detect what color the different parts of the fish's body are.

The first part of the process is to run the 2-D DWT on each of the color matricies. It is run 3 times so that the resulting picture is 1/8 of the resolution of the original matricies, and there are only high values where the color is relatively constant for a large area. This essentially provides a method for Low-Pass filtering the picture and finding only large blocks of color.

Next, the picture is filtered by dropping any values that are lower than a threshold and setting any values over that threshold to 1. This drops all areas in the picture that are not very intense, or where the values are not constant for a large area. Now, the picture has only ones whereever the large blocks of color are.

The next step is to count all the different blocks of ones, which is done by using the Matlab command, bwlabel. Next, each block is examined one by one to see what size it is, and where in the picture its located. From this, it can be determined what color the body, head, and tail of the fish is. If they match the pattern for either type of fish, then the test classifies it as that type. Because this is the hardest test to satisfy, it is also the most heavily weighted test in the entire process.

## Example 1: Code for Feature Detection

% This function takes in a 3D image matrix of three colors, red, green, and blue and uses the
% 2-D DWT to low poss filter them and decrease their resolution.  It then looks for blocks of
% color and outputs a matrix with the size, and location of each of the different features of
% each color.  These can then be analyzed to see if they show evidence of a specific fish type.

function [rfeats,gfeats,bfeats] = featuredet(image)

redimage = image(:,:,1);
greenimage = image(:,:,2);
blueimage = image(:,:,3);

rfeats = [0 0 0 0 0 0 0 0];
gfeats = [0 0 0 0 0 0 0 0];
bfeats = [0 0 0 0 0 0 0 0];

% Run the 2-D DWT on the different colors to reduce the resolution of the picture and
% effectively low-pass filter the image.
dwtr = dwt2(redimage, 'haar');
dwtr2 = dwt2(dwtr, 'haar');
dwtr3 = dwt2(dwtr2, 'haar');
%dwtr3feat = (dwtr3 > 6)
dwtr4 = dwt2(dwtr3, 'haar');
%dwtr4feat = (dwtr4 > 13);

dwtg = dwt2(greenimage, 'haar');
dwtg2 = dwt2(dwtg, 'haar');
dwtg3 = dwt2(dwtg2, 'haar');
dwtg4 = dwt2(dwtg3, 'haar');

dwtb = dwt2(blueimage, 'haar');
dwtb2 = dwt2(dwtb, 'haar');
dwtb3 = dwt2(dwtb2, 'haar');
dwtb4 = dwt2(dwtb3, 'haar');

% Set everything below a threshold to 0 and everything above to 1 and then
% number every group of ones in the binary image
[redfeatures, numred] = bwlabel(dwtr3>5);
[greenfeatures, numgreen] = bwlabel(dwtg3>5);
[bluefeatures, numblue] = bwlabel(dwtb3>5);

% Cycle through each different feature and find its location and size
for a = 1:numred
rowval = sum(redfeatures==a);
colval = sum((redfeatures==a)')';
sizeval = size(redfeatures);
j = 1;
left = 0;
while rowval(j)<1
left = j;
j = j+1;
end
j = 1;
right = sizeval(2);
while rowval(sizeval(2)-j+1)<1
right = sizeval(2)-j+1;
j = j+1;
end
j = 1;
top = 0;
while colval(j)<1
top = j;
j = j+1;
end
j = 1;
bottom = sizeval(1);
while colval(sizeval(1)-j+1)<1
bottom = sizeval(1)-j+1;
j = j+1;
end
sumval = sum(rowval);

rfeats(a,:) = [top bottom bottom-top left right right-left (right-left)./(bottom-top) sumval];
end

for b = 1:numgreen
rowval = sum(greenfeatures==b);
colval = sum((greenfeatures==b)')';
sizeval = size(greenfeatures);
j = 1;
while rowval(j)<1
left = j;
j = j+1;
end
j = 1;
while rowval(sizeval(2)-j+1)<1
right = sizeval(2)-j+1;
j = j+1;
end
j = 1;
while colval(j)<1
top = j;
j = j+1;
end
j = 1;
while colval(sizeval(1)-j+1)<1
bottom = sizeval(1)-j+1;
j = j+1;
end
sumval = sum(rowval);
gfeats(b,:) = [top bottom bottom-top left right right-left (right-left)./(bottom-top) sumval];
end

for c = 1:numblue
rowval = sum(bluefeatures==c);
colval = sum((bluefeatures==c)')';
sizeval = size(bluefeatures);
j = 1;
while rowval(j)<1
left = j;
j = j+1;
end
j = 1;
while rowval(sizeval(2)-j+1)<1
right = sizeval(2)-j+1;
j = j+1;
end
j = 1;
while colval(j)<1
top = j;
j = j+1;
end
j = 1;
while colval(sizeval(1)-j+1)<1
bottom = sizeval(1)-j+1;
j = j+1;
end
sumval = sum(rowval);
bfeats(c,:) = [top bottom bottom-top left right right-left (right-left)./(bottom-top) sumval];
end


## Example 2: Code for Feature Analysis

% This function takes a 3D image and runs the feature detector on it, which gives matricies
% containing the sizes and shapes of the different features.  It then decides what color the
% fish's body, head, and tail are, or whether the can't be determined by the features.

[rfeats,gfeats,bfeats] = featuredet(fishimage);

impfeats = [0 0 0; 0 0 0; 0 0 0];

% This section takes each feature located by the feature detector and decides if they
% are evidence or a body, head, or tail of the fish being that color.
for a = 1:size(rfeats,1)
% If the feature is extremely long, it is a body
if and(rfeats(a,6)>30, rfeats(a,8)>30)
impfeats(1,1) = 1;
end
% If the feature is far to the right, it is a head
if and(rfeats(a,4)>38, rfeats(a,8)>20)
impfeats(1,2) = 1;
end
% If the features is far to the left, it is a tail
if and(rfeats(a,5)<25, rfeats(a,8)>6)
impfeats(1,3) = 1;
end
end

for b = 1:size(gfeats,1)
if and(gfeats(b,6)>30, gfeats(b,8)>30)
impfeats(2,1) = 1;
end
if and(gfeats(b,4)>38, gfeats(b,8)>10)
impfeats(2,2) = 1;
end
if and(gfeats(b,5)<25, gfeats(b,8)>6)
impfeats(2,3) = 1;
end
end

for c = 1:size(bfeats,1)
if and(bfeats(c,6)>30, bfeats(c,8)>30)
impfeats(3,1) = 1;
end
if and(bfeats(c,4)>38, bfeats(c,8)>10)
impfeats(3,2) = 1;
end
if and(bfeats(c,5)<25, bfeats(c,8)>6)
impfeats(3,3) = 1;
end
end

% This section looks at each of the columns of the feature matrix and then
% outputs which color pattern they are.
if and(impfeats(1,1) == 1, and(impfeats (2,1) == 1, impfeats (3,1) == 1))
body = 'rgb';
end
if and(impfeats(1,1) == 1, and(impfeats (2,1) == 1, impfeats (3,1) == 0))
body = 'rg ';
end
if and(impfeats(1,1) == 1, and(impfeats (2,1) == 0, impfeats (3,1) == 1))
body = 'rb ';
end
if and(impfeats(1,1) == 1, and(impfeats (2,1) == 0, impfeats (3,1) == 0))
body = 'r  ';
end
if and(impfeats(1,1) == 0, and(impfeats (2,1) == 1, impfeats (3,1) == 1))
body = 'gb ';
end
if and(impfeats(1,1) == 0, and(impfeats (2,1) == 1, impfeats (3,1) == 0))
body = 'g  ';
end
if and(impfeats(1,1) == 0, and(impfeats (2,1) == 0, impfeats (3,1) == 1))
body = 'b  ';
end
if and(impfeats(1,1) == 0, and(impfeats (2,1) == 0, impfeats (3,1) == 0))
body = 'cbd';
end

if and(impfeats(1,2) == 1, and(impfeats (2,2) == 1, impfeats (3,2) == 1))
end
if and(impfeats(1,2) == 1, and(impfeats (2,2) == 1, impfeats (3,2) == 0))
end
if and(impfeats(1,2) == 1, and(impfeats (2,2) == 0, impfeats (3,2) == 1))
end
if and(impfeats(1,2) == 1, and(impfeats (2,2) == 0, impfeats (3,2) == 0))
end
if and(impfeats(1,2) == 0, and(impfeats (2,2) == 1, impfeats (3,2) == 1))
end
if and(impfeats(1,2) == 0, and(impfeats (2,2) == 1, impfeats (3,2) == 0))
end
if and(impfeats(1,2) == 0, and(impfeats (2,2) == 0, impfeats (3,2) == 1))
end
if and(impfeats(1,2) == 0, and(impfeats (2,2) == 0, impfeats (3,2) == 0))
end

if and(impfeats(1,3) == 1, and(impfeats (2,3) == 1, impfeats (3,3) == 1))
tail = 'rgb';
end
if and(impfeats(1,3) == 1, and(impfeats (2,3) == 1, impfeats (3,3) == 0))
tail = 'rg ';
end
if and(impfeats(1,3) == 1, and(impfeats (2,3) == 0, impfeats (3,3) == 1))
tail = 'rb ';
end
if and(impfeats(1,3) == 1, and(impfeats (2,3) == 0, impfeats (3,3) == 0))
tail = 'r  ';
end
if and(impfeats(1,3) == 0, and(impfeats (2,3) == 1, impfeats (3,3) == 1))
tail = 'gb ';
end
if and(impfeats(1,3) == 0, and(impfeats (2,3) == 1, impfeats (3,3) == 0))
tail = 'g  ';
end
if and(impfeats(1,3) == 0, and(impfeats (2,3) == 0, impfeats (3,3) == 1))
tail = 'b  ';
end
if and(impfeats(1,3) == 0, and(impfeats (2,3) == 0, impfeats (3,3) == 0))
tail = 'cbd';
end


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