求matlab彩色图片的颜色特征提取算法的代码，和纹理特征提取的代码。传统方法即可。
发布网友
发布时间：2022-04-26 18:27
共2个回答
热心网友
时间：2023-10-21 00:08
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
热心网友
时间：2023-10-21 00:09
你好，不知道这是什么时候提的问题了，我用的matlab的颜色纹理解决行人重识别问题。方法是LOMO+XQDA，这是项目主页网页链接希望可以帮到你
