Figure 1. |
In MATLAB, HSV color space of an image is three dimensional matrix and each matrix represents each of the 3 component (Hue,Saturation,Value). Hue and saturation range between zero and one. While saturation defines colorfulness hue is specific to the color.
%MATLAB CODE:
A = imread('swimmer.jpg');
figure,imshow(A);title('Original
Image');
Original Image |
HSV = rgb2hsv(A);
H = HSV(:,:,1); %Hue
figure,imshow(H);colorbar;
Hue |
H( H > mean2(H) ) = 1;
HSV(:,:,1) = H;
C = hsv2rgb(HSV);
figure,imshow(C);title('Hue Modified');
Hue Modified |
EXPLANATION:
The original image is in RGB format and it is converted to HSV color space
using the MATLAB command ‘rgb2hsv’. The resultant is a three dimensional matrix
with Hue, Saturation and Value components in each one of them. By comparing the
original RGB image and the Hue component, we can understand that the blue color
as high value comparing to other colors in the original image.
Hue is a color wheel, where the
colors start from red, then move on to yellow, green, cyan, blue, magenta and ends
up again in red.
In our example,the values above the average in
the Hue matrix is made 1.(i.e. red). The background of the image is changed
from blue to red.
If the whole image including the swimmer needs to
be changed to red then instead of finding the average or masking the image,
assign zero or 1 to Hue matrix.
After the modification, use
‘hsv2rgb’ command to return back to RGB color space.
%MATLAB CODE:
HSV = rgb2hsv(A);
S = HSV(:,:,2); %Saturation
figure,imshow(S);colorbar;
Saturation |
S(:,:)=0;
HSV(:,:,2) = S;
C = hsv2rgb(HSV);
figure,imshow(C);title('Saturation
Modified');
Saturation Modified |
EXPLANATION:
In this example, the saturation component is modified. The high values
of Saturation illustrates that the regions are bright and colorful while the
low values illustrates that they are dull and colorless.When saturation
matrix is made zero, the colorfulness is completely lost. The above figure
clearly shows the gray shade image that is obtained as a
result of modifying the saturation matrix.
%MATLAB CODE:
HSV = rgb2hsv(A);
H = HSV(:,:,1); %Hue
S = HSV(:,:,2); %Saturation
H( H > mean2(H) ) = 0.42;
S( H < mean2(H) )=0;
S( H >= mean2(H) )=1;
HSV(:,:,2) = S;
HSV(:,:,1) = H;
C = hsv2rgb(HSV);
figure,imshow(C);title('Saturation
Modified - Background');
EXPLANATION:
In this example, both the Hue and
the Saturation matrices are modified simultaneously. The background color is
changed from blue to green after changing the Hue matrix. The Saturation matrix
is changed partially such that the background is not affected but the color on
the swimmer is made gray.
%MATLAB CODE:
HSV = rgb2hsv(A);
H = HSV(:,:,1); %Hue
S = HSV(:,:,2); %Saturation
S( H < mean2(H) )=1;
S( H >= mean2(H) )=0;
HSV(:,:,2) = S;
HSV(:,:,1) = H;
C = hsv2rgb(HSV);
figure,imshow(C);title('Saturation
Modified - Foreground');
EXPLANATION:
In this example, the background
color is made shades of gray while the swimmer still retains the color. The
masking is done based on the foreground and background on the saturation matrix
and a value of zero is assigned to the background and one is assigned to
foreground. From this example, it is evident that if the saturation matrix contains
zero then the image in RGB color space will contain shades of gray whereas if
the saturation matrix contains one then it will contain fully saturated more
colorful image in the RGB color space.
The image(Figure.1) above shows the swimmer and different background
colors by modifying the Hue matrix.