其实学数字图像处理,关键的不是源代码(和一般编程还是有区别的,这个是经验之谈,其实一般博导未必会编程,但是你和他说说你的方法,他一般都能切中要害),而是你能理解基于概念及适用场所。
基于颜色、纹理、形状都属于低层特征,这些你理解就够了,关键是对你的课题适合哪种方法来映射到高层语义上面,例如:识别物体轮廓,那可能形状就比较适合等。
我之所以写上面那段话,主要是我感觉你索取代码也不说明具体要求,也就是方向不明确。
如今颜色特征提取算法有很多,诸如颜色直方图、颜色矩、颜色集、颜色聚合向量、颜色相关图等,既然你没说,我就给个IEEE CSVT 2001的一篇关于颜色直方图法的论文(源码版权归作者所有):
function colorhist = colorhist(rgb)
% CBIR_colorhist() --- color histogram calculation
% input: MxNx3 image data, in RGB
% output: 1x256 colorhistogram == (HxSxV = 16x4x4)
% as the MPEG-7 generic color histogram descriptor
% [Ref] Manjunath, B。
S。; Ohm, J。-R。; Vasudevan, V。V。; Yamada, A。, "Color and texture descriptors"
% IEEE Trans。 CSVT, Volume: 11 Issue: 6 , Page(s): 703 -715, June 2001 (section III。
B)
% check input
if size(rgb,3)~=3
error('3 components is needed for histogram');
end
% globals
H_BITS = 4; S_BITS = 2; V_BITS = 2;
%rgb2hsv可用rgb2hsi代替,见你以前的提问。
hsv = uint8(255*rgb2hsv(rgb));
imgsize = size(hsv);
% get rid of irrelevant boundaries
i0=round(0。
05*imgsize(1)); i1=round(0。95*imgsize(1));
j0=round(0。05*imgsize(2)); j1=round(0。95*imgsize(2));
hsv = hsv(i0:i1, j0:j1, :);
% histogram
for i = 1 : 2^H_BITS
for j = 1 : 2^S_BITS
for k = 1 : 2^V_BITS
colorhist(i,j,k) = sum(sum( 。
。。
bitshift(hsv(:,:,1),-(8-H_BITS))==i-1 &。。。
bitshift(hsv(:,:,2),-(8-S_BITS))==j-1 &。。。
bitshift(hsv(:,:,3),-(8-V_BITS))==k-1 ));
end
end
end
colorhist = reshape(colorhist, 1, 2^(H_BITS S_BITS V_BITS));
% normalize
colorhist = colorhist/sum(colorhist);
%基于纹理特征提取灰度共生矩阵用于纹理判断
% Calculates cooccurrence matrix
% for a given direction and distance
%
% out = cooccurrence (input, dir, dist, symmetric);
%
% INPUT:
% input: input matrix of any size
%
% dir: direction of evaluation
% "dir" value Angle
% 0 0
% 1 -45
% 2 -90
% 3 -135
% 4 -180
% 5 135
% 6 90
% 7 45
%
% dist: distance between pixels
%
% symmetric: 1 for symmetric version
% 0 for non-symmetric version
%
% eg: out = cooccurrence (input, 0, 1, 1);
% Author: Baran Aydogan (15。
07。2006)
% RGI, Tampere University of Technology
% baran。aydogan@tut。
fi
function out = cooccurrence (input, dir, dist, symmetric);
input = round(input);
[r c] = size(input);
min_intensity = min(min(input));
max_intensity = max(max(input));
out = zeros(max_intensity-min_intensity 1);
if (dir == 0)
dir_x = 0; dir_y = 1;
end
if (dir == 1)
dir_x = 1; dir_y = 1;
end
if (dir == 2)
dir_x = 1; dir_y = 0;
end
if (dir == 3)
dir_x = 1; dir_y = -1;
end
if (dir == 4)
dir_x = 0; dir_y = -1;
end
if (dir == 5)
dir_x = -1; dir_y = -1;
end
if (dir == 6)
dir_x = -1; dir_y = 0;
end
if (dir == 7)
dir_x = -1; dir_y = 1;
end
dir_x = dir_x*dist;
dir_y = dir_y*dist;
out_ind_x = 0;
out_ind_y = 0;
for intensity1 = min_intensity:max_intensity
out_ind_x = out_ind_x 1;
out_ind_y = 0;
[ind_x1 ind_y1] = find (input == intensity1);
ind_x1 = ind_x1 dir_x;
ind_y1 = ind_y1 dir_y;
for intensity2 = min_intensity:max_intensity
out_ind_y = out_ind_y 1;
[ind_x2 ind_y2] = find (input == intensity2);
count = 0;
for i = 1:size(ind_x1,1)
for j = 1:size(ind_x2,1)
if ( (ind_x1(i) == ind_x2(j)) && (ind_y1(i) == ind_y2(j)) )
count = count 1;
end
end
end
out(out_ind_x, out_ind_y) = count;
end
end
if (symmetric)
if (dir > 4)
dir = dir 4;
else
dir = mod(dir,4);
end
out = out cooccurrence (input, dir, dist, 0);
end。
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