LBP: local Binary Patterns翻译成中文就是局部二进模式,是图像的一个很重要的纹理特征(Texture Feature)。
这个特征的提出当然要看最早由Timo Ojala在2002年发表的那篇PAMI(听到这个杂志就知道这文章就有多牛了),即 Multiresolution Gray-scale and Rotation Invariant Texture Classification with Local Binary Patterns.
有关LBP的具体介绍可看那篇paper,也可以看这个链接,click.
它的matlab实现当然要直接到它的作者那里,那当然是最权威的实现方式了。不过有通力的人也可以自己写一个,反正也来难。
下载请点击如下:LBP Matlab implement download.
注:这种特征可用来作CBIR (Content-based Image Retrieval)或image reranking.
lbp.m 文件如下:(原文件下载地址:link)
%LBP returns the local binary pattern image or LBP histogram of an image. % J = LBP(I,R,N,MAPPING,MODE) returns either a local binary pattern % coded image or the local binary pattern histogram of an intensity % image I. The LBP codes are computed using N sampling points on a % circle of radius R and using mapping table defined by MAPPING. % See the getmapping function for different mappings and use 0 for % no mapping. Possible values for MODE are % 'h' or 'hist' to get a histogram of LBP codes % 'nh' to get a normalized histogram % Otherwise an LBP code image is returned. % % J = LBP(I) returns the original (basic) LBP histogram of image I % % J = LBP(I,SP,MAPPING,MODE) computes the LBP codes using n sampling % points defined in (n * 2) matrix SP. The sampling points should be % defined around the origin (coordinates (0,0)). % % Examples % -------- % I=imread('rice.png'); % mapping=getmapping(8,'u2'); % H1=LBP(I,1,8,mapping,'h'); %LBP histogram in (8,1) neighborhood % %using uniform patterns % subplot(2,1,1),stem(H1); % % H2=LBP(I); % subplot(2,1,2),stem(H2); % % SP=[-1 -1; -1 0; -1 1; 0 -1; -0 1; 1 -1; 1 0; 1 1]; % I2=LBP(I,SP,0,'i'); %LBP code image using sampling points in SP % %and no mapping. Now H2 is equal to histogram % %of I2. function result = lbp(varargin) % image,radius,neighbors,mapping,mode) % Version 0.3.2 % Authors: Marko Heikkil�and Timo Ahonen % Changelog % Version 0.3.2: A bug fix to enable using mappings together with a % predefined spoints array % Version 0.3.1: Changed MAPPING input to be a struct containing the mapping % table and the number of bins to make the function run faster with high number % of sampling points. Lauge Sorensen is acknowledged for spotting this problem. % Check number of input arguments. error(nargchk(1,5,nargin)); image=varargin{1}; d_image=double(image); if nargin==1 spoints=[-1 -1; -1 0; -1 1; 0 -1; -0 1; 1 -1; 1 0; 1 1]; neighbors=8; mapping=0; mode='h'; end if (nargin == 2) && (length(varargin{2}) == 1) error('Input arguments'); end if (nargin > 2) && (length(varargin{2}) == 1) radius=varargin{2}; neighbors=varargin{3}; spoints=zeros(neighbors,2); % Angle step. a = 2*pi/neighbors; for i = 1:neighbors spoints(i,1) = -radius*sin((i-1)*a); spoints(i,2) = radius*cos((i-1)*a); end if(nargin >= 4) mapping=varargin{4}; if(isstruct(mapping) && mapping.samples ~= neighbors) error('Incompatible mapping'); end else mapping=0; end if(nargin >= 5) mode=varargin{5}; else mode='h'; end end if (nargin > 1) && (length(varargin{2}) > 1) spoints=varargin{2}; neighbors=size(spoints,1); if(nargin >= 3) mapping=varargin{3}; if(isstruct(mapping) && mapping.samples ~= neighbors) error('Incompatible mapping'); end else mapping=0; end if(nargin >= 4) mode=varargin{4}; else mode='h'; end end % Determine the dimensions of the input image. [ysize xsize] = size(image); miny=min(spoints(:,1)); maxy=max(spoints(:,1)); minx=min(spoints(:,2)); maxx=max(spoints(:,2)); % Block size, each LBP code is computed within a block of size bsizey*bsizex bsizey=ceil(max(maxy,0))-floor(min(miny,0))+1; bsizex=ceil(max(maxx,0))-floor(min(minx,0))+1; % Coordinates of origin (0,0) in the block origy=1-floor(min(miny,0)); origx=1-floor(min(minx,0)); % Minimum allowed size for the input image depends % on the radius of the used LBP operator. if(xsize < bsizex || ysize < bsizey) error('Too small input image. Should be at least (2*radius+1) x (2*radius+1)'); end % Calculate dx and dy; dx = xsize - bsizex; dy = ysize - bsizey; % Fill the center pixel matrix C. C = image(origy:origy+dy,origx:origx+dx); d_C = double(C); bins = 2^neighbors; % Initialize the result matrix with zeros. result=zeros(dy+1,dx+1); %Compute the LBP code image for i = 1:neighbors y = spoints(i,1)+origy; x = spoints(i,2)+origx; % Calculate floors, ceils and rounds for the x and y. fy = floor(y); cy = ceil(y); ry = round(y); fx = floor(x); cx = ceil(x); rx = round(x); % Check if interpolation is needed. if (abs(x - rx) < 1e-6) && (abs(y - ry) < 1e-6) % Interpolation is not needed, use original datatypes N = image(ry:ry+dy,rx:rx+dx); D = N >= C; else % Interpolation needed, use double type images ty = y - fy; tx = x - fx; % Calculate the interpolation weights. w1 = (1 - tx) * (1 - ty); w2 = tx * (1 - ty); w3 = (1 - tx) * ty ; w4 = tx * ty ; % Compute interpolated pixel values N = w1*d_image(fy:fy+dy,fx:fx+dx) + w2*d_image(fy:fy+dy,cx:cx+dx) + ... w3*d_image(cy:cy+dy,fx:fx+dx) + w4*d_image(cy:cy+dy,cx:cx+dx); D = N >= d_C; end % Update the result matrix. v = 2^(i-1); result = result + v*D; end %Apply mapping if it is defined if isstruct(mapping) bins = mapping.num; for i = 1:size(result,1) for j = 1:size(result,2) result(i,j) = mapping.table(result(i,j)+1); end end end if (strcmp(mode,'h') || strcmp(mode,'hist') || strcmp(mode,'nh')) % Return with LBP histogram if mode equals 'hist'. result=hist(result(:),0:(bins-1)); if (strcmp(mode,'nh')) result=result/sum(result); end else %Otherwise return a matrix of unsigned integers if ((bins-1)<=intmax('uint8')) result=uint8(result); elseif ((bins-1)<=intmax('uint16')) result=uint16(result); else result=uint32(result); end end end